An important factor to notice here is that the reactions are similar since in all of them, two reactants are producing one product, and all the coefficients are one in the balanced equation.

(a) The reaction, A + D ⇆ AD has the largest equilibrium constant because it produces the most product at equilibrium.

(b) The reaction, A + C ⇆ AC has the smallest equilibrium constant because it produces the amount of product at equilibrium.

a) Initially, hydrogen has the highest concentration judging from the number of spheres. Notice that blue spheres correspond to hydrogen since they are smaller atoms than those of nitrogen. Therefore, the blue line corresponds to the concentration of H₂ which is decreasing with time.

The concentration of N₂ is shown by the green line because it is decreasing as the reaction continues.

Ammonia, on the other hand, is not present in the beginning, and therefore, the red line represents its concentration which is increasing with time.

b) Let’s first write the balanced chemical equation for this reaction:

3H₂(g) + N₂(g) ⇆ 2NH₃(g)

When a reaction starts, only H₂ and N₂ are present in the vessel and therefore, their concentrations decrease while the concentration of ammonia increases with time. Another observation is the difference in the concentration change of hydrogen and nitrogen.  Because the stoichiometry in the balanced equation dictates that for every 1 molecule of N₂, 3 molecules of H₂ are needed, the H₂ concentration initially decreases more rapidly as compared to the initial decrease in N₂ concentration.

Likewise, the NH₃ plot initially increases faster than the N₂ plot decreases since, here as well, the ratio of these two gases is 2 : 1 meaning there is 2 times more ammonia produced than nitrogen is consumed at any time of reaction.

c) The equilibrium is reached when the concentrations of reactants and products do not change anymore. This can be seen when the concentration lines hit a plateau at about 4-hour mark.

The equilibrium constant is the ratio of the equilibrium concentrations of the products and reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

\[K = \frac{{{\rm{[}}{{\rm{N}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{\rm{O}}{{\rm{]}}^{\rm{2}}}}}{{{{{\rm{[NO]}}}^{\rm{2}}}{{{\rm{[}}{{\rm{H}}_{\rm{2}}}{\rm{]}}}^{\rm{2}}}}}\]

\[K = \frac{{{\rm{(}}4.8 \times {{10}^{ – 2}}{\rm{)(}}2.4 \times {{10}^{ – 2}}{{\rm{)}}^{\rm{2}}}}}{{{{(3.2 \times {{10}^{ – 3}})}^{\rm{2}}}{{(6.7 \times {{10}^{ – 6}})}^{\rm{2}}}}} = 6.01 \times {10^{10}}\]

\[K = \frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{OH][}}{{\rm{H}}_{\rm{2}}}{\rm{O]}}}}{{{\rm{[C}}{{\rm{O}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{{\rm{]}}^{\rm{3}}}}}\]

\[K = \frac{{{\rm{(}}4.7 \times {{10}^2}{\rm{)(}}5.7 \times {{10}^4}{\rm{)}}}}{{{\rm{(}}0.061{\rm{)(}}0.079{{\rm{)}}^{\rm{3}}}}} = 8.9 \times {10^{11}}\]

Remember, when a reaction is reversed, then Knew = 1/Koriginal. When a reaction is multiplied by a value of n, then Knew = (Koriginal)ⁿ.

a) Comparing the equation CH₄(g) ⇆ 1/2C₂H₂(g) + 3/2H₂(g) to the original, we see that it is multiplied by 1/2. Therefore, Knew = (Koriginal)ⁿ = (0.154)^1/2= 0.392

b) Comparing the equation 4CH₄(g) ⇆ 2C₂H₂(g) + 6H₂(g) to the original, we see that it is multiplied by 2. Therefore, Knew = (Koriginal)ⁿ = (0.154)²= 0.0237

c) Comparing the equation 6CH₄(g) ⇆ 3C₂H₂(g) + 9H₂(g) to the original, we see that it is multiplied by 3. Therefore, Knew = (Koriginal)ⁿ = (0.154)³= 0.00365

Remember, when a reaction is reversed, then Knew = 1/Koriginal. When a reaction is multiplied by a value of n, then Knew = (Koriginal)ⁿ.

a) Comparing the equation CO (g) + 2H₂(g) ⇆ CH₃OH(g) to the original, we see that it is reversed. Therefore, Knew = 1/Koriginal= 1/4.42 x 10^-5= 22,624.43 = 2.26 x 10⁴

b) Comparing the equation 2CH₃OH(g) ⇆ 2CO (g) + 4H₂(g) to the original, we see that it is multiplied by 2. Therefore, Knew = (Koriginal)ⁿ = (4.42 x 10^-5)²= 1.96 x 10^-9

c) Comparing the equation 1/2CH₃OH(g) ⇆ 1/2CO (g) + H₂(g) to the original, we see that it is multiplied by 1/2. Therefore, Knew = (Koriginal)ⁿ = (4.42 x 10^-5)^1/2= 0.00665=6.65 x 10^-3

There are two SO₂ reactant molecules in the target reaction, so reverse the first reaction and multiply by 2:

S(s) + O₂(g) ⇆ SO₂(g) ⇒ 2SO₂(g) ⇆ 2S(s) + 2O₂(g)

The equilibrium constant for this reaction is:

K^r_a = (1/k_a)² = (1/3.2 x 10⁴⁵)² = 9.77 x 10^-92

Now, add the two reactions to make sure it matches the target equation:

2S(s) + 3O₂(g) ⇆ 2SO₃(g)

2SO₂(g) ⇆ 2S(s) + 2O₂(g)
__________________

2S(s) + 3O₂(g) + 2SO₂(g) ⇆ 2SO₃(g) + 2S(s) + 2O₂(g)

O₂(g) + 2SO₂(g) ⇆ 2SO₃(g)

The equations add up, so the last part is to calculate the equilibrium constant.

K_c = K^r_a x K_b = 9.77 x 10^-92 x 3.2 x 10¹²⁴ = 3.1 x 10³³

There are 1/2 moles of N₂ reactant molecules in the target reaction, so reverse the first reaction and multiply by 1/2:

2NO(g) ⇆ N₂(g) + O₂(g) ⇒ 1/2N₂(g) + 1/2O₂(g) ⇆ NO(g)

The equilibrium constant for this reaction is:

K^r_a = (1/k_a)^1/2 = (1/3.8 x 10³²)^1/2 = 5.13 x 10^-17

Now, add the two reactions to make sure it matches the target equation:

1/2N₂(g) + 1/2O₂(g) ⇆ NO(g)

NO(g) + 1/2Cl₂(g) ⇆ NOCl(g)

_____________________

1/2N₂(g) + 1/2O₂(g) + NO(g) + 1/2Cl₂(g) ⇆ NO(g) + NOCl(g)

1/2N₂(g) + 1/2O₂(g) + 1/2Cl₂(g) ⇆ NOCl(g)

The equations add up, so the last part is to calculate the equilibrium constant.

K_c = K^r_a x K_b = 5.13 x 10^-17 x 6.4 = 3.28 x 10^-16

\[{K_p} = \frac{{{P^2}_{{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}{{{P^3}_{{{\rm{H}}_{\rm{2}}}}{P_{{{\rm{N}}_{\rm{2}}}}}}}\]

\[{K_p} = \frac{{{{(5.2 \times {{10}^9})}^2}}}{{{{(4.7 \times {{10}^3})}^3}(6.1 \times {{10}^2})}} = 4.3 \times {10^5}\]

Kp = K(RT)^Δn

Where Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. In the first reaction, Δn=0, therefore, Kp = K.

Kp = Kc(RT)Δn

Where Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. For this reaction, Δn=1+2-1=2.

Next, we need to determine the Kc and use that value to calculate the Kp.

The expression for Kc is:

\[K = \frac{{[{\rm{CO}}]{{[{{\rm{H}}_{\rm{2}}}]}^{\rm{2}}}}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}]}}\]

Therefore,

\[K = \frac{{(0.350){{({\rm{1}}{\rm{.65}})}^2}}}{{{\rm{(0}}{\rm{.240)}}}} = 3.97\]

K_p = K(RT)² = 3.97(0.08206 L atm/K•mol × 300. K)2 = 2,406 = 2.41 x 103

Kp = K(RT)^Δn

Where Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.

a) Δn = 2-1 = 1

Kp = K(RT) = 4.6 x 10^-4(0.08206 L atm/K•mol × 298 K) = 1.1 x 10^-2

b) Δn = 2- (3+ 1) = -2

Kp = K(RT) = 6.7 x 10⁹(0.08206 L atm/K•mol × 298 K)^-2 = 1.6 x 10¹¹

c) Δn = 2- (1+ 1) = 0

Kp = K(RT)⁰ = K = 5.20 x 10¹⁸

Kp = Kc(RT)^Δn , therefore, Kc = Kp/(RT)^Δn

a) Δn = 1-(1+1) = -1

Kc = 5.3 x 10⁶/(0.08206 L atm/K•mol × 298 K)^-1 = 129605 = 1.3 x 10⁵

b) Δn = 1+3-(1+1) = 2

Kc = 7.7 x 10⁸/(0.08206 L atm/K•mol × 298 K)² = 31487 =3.1 x 10⁴

c) Δn = 2-(1+2) = -1

Kc = 6.26 x 10⁵/(0.08206 L atm/K•mol × 298 K)^-1 = 15308 = 1.5 x 10⁴

1) Determine the equilibrium constant using the concentration of the reactants and products in the table (row 1).

\[K = \frac{{{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}^{\rm{2}}}}}{{{\rm{[}}{{\rm{N}}_{\rm{2}}}{\rm{][}}{{\rm{H}}_{\rm{2}}}{{\rm{]}}^{\rm{3}}}}}\]

\[K = \frac{{{{{\rm{[0}}{\rm{.518]}}}^{\rm{2}}}}}{{{\rm{[0}}{\rm{.142][0}}{\rm{.129}}{{\rm{]}}^{\rm{3}}}}} = 8,80 \times {10^2}\]

2) Assign the concentration of ammonia as x, and set up an equation based on the equilibrium constant.

\[K = \frac{{{{{\rm{(X)}}}^{\rm{2}}}}}{{{\rm{(0}}{\rm{.125)(0}}{\rm{.116}}{{\rm{)}}^{\rm{3}}}}} = 7.8\]

X² = 0.001521874, X = 0.0390 = 3.9 x 10^-2

3) Assign the concentration of hydrogen as x, and set up an equation based on the equilibrium constant.

\[K = \frac{{{{{\rm{(0}}{\rm{.712)}}}^{\rm{2}}}}}{{{\rm{(0}}{\rm{.136)(X}}{{\rm{)}}^{\rm{3}}}}} = 0.0647\]

0.0087992 X³ = 0.506944

X = 3.86

Remember, when converting Kc and Kp, the value for R is most often used as 0.08206 L atm/ mol K, and therefore, the units for pressure should also be in atm.

\[{P_{({\rm{NO}})}}{\rm{ = }}\;\;{\rm{128torr}}\;\;{\rm{x}}\;\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}\;{\rm{ = }}\;\;{\rm{0}}.{\rm{168}}\;{\rm{atm}}\]

\[{P_{({\rm{C}}{{\rm{l}}_{\rm{2}}})}}\; = \;{\rm{176}}\;{\rm{torr\;}}\; \times \;{\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{tor}}}}\,{\rm{\; = \;}}\,{\rm{0}}.{\rm{232}}\,{\rm{atm}}\]

Next, use the Kp expression to assign the numbers and an unknown for the P_(NOCl):

\[{K_p} = \frac{{{P^2}_{{\rm{NOCl}}}}}{{{P^2}_{{\rm{NO}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\]

\[{K_p} = \frac{{{{\rm{X}}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.168}}} \right)}^{\rm{2}}}{\rm{0}}{\rm{.232}}}}\;{\rm{ = }}\;{\rm{31}}{\rm{.6}}\]

X = 0.455 atm = 346 torr

Check:

\[{K_p} = \frac{{{{\rm{0.455}}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.168}}} \right)}^{\rm{2}}}{\rm{0}}{\rm{.232}}}}\;{\rm{ = }}\;{\rm{31}}{\rm{.6}}\]

✔

The concentrations of the gases are going to be moles over the container volume (10.0 L). So, [CO] = 0.0400 mol/L and [H₂O] = 0.0500 mol/L.

Let’s set up an ICE table assigning x mol/L as the depletion in the concentration of either CO or H₂O. The increase of H₂ and CO₂ concentration is going to be x which is their equilibrium concentration. The equilibrium concentration of CO and H₂O is going to be 0.0400-x and 0.0500-x respectively.

H₂O(g) + CO(g) ⇆ H₂(g) + CO₂(g)

The equilibrium constat is:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]\left[ {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right]}}{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]\left[ {{\rm{CO}}} \right]}}\; = \,0.495\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left( {\rm{x}} \right)\left( {\rm{x}} \right)}}{{\left( {{\rm{0}}{\rm{.0500 – x}}} \right)\left( {{\rm{0}}{\rm{.0400 – x}}} \right)}}\;{\rm{ = }}\,{\rm{0}}{\rm{.495}}\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{\left( {{\rm{0}}.{\rm{0500 – x}}} \right)\left( {{\rm{0}}.{\rm{0400 – x}}} \right)}}\;{\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{495}}\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}\; – \;0.09x\; + \;0.002}}\;{\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{495}}\]

0.505x² + 0.04455x – 0.00099 = 0 

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]

a = 0.505, b = 0.04455, c = -0.00099

x = 0.0184 mol/L

The last step is to convert the concentration to the number of moles by multiplying by 10 since the reaction is carried out in a 10.0 L container. 

n (H₂) = 0.0184 mol/L x 10.0 L = 0.184 mol

First, determine the initial concentration of HI.

M (HI) = 6.25 mol ÷ 15.0 L = 0.417 mol/L

Then, set up the ICE table, assuming x mol/L of HI decomposed to form H₂ and I_2. From the stoichiometric ratio, we know that 0.5x mol/l of H₂ and I₂ will be formed:

H₂(g) + I₂(g) ⇆ 2HI(g)

The equilibrium constant is:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left[ {{\rm{HI}}} \right]}^2}}}{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]\left[ {{{\rm{I}}_{\rm{2}}}} \right]}}\; = \,3.74{\rm{ }}x{\rm{ }}{10^5}\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}^2}}}{{\left( {{\rm{0}}{\rm{.5x}}} \right)\left( {{\rm{0}}{\rm{.5x}}} \right)}}\; = \,3.74{\rm{ }} \times {\rm{ }}{10^5}\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}^{\rm{2}}}}}{{{\rm{0}}{\rm{.25}}{{\rm{x}}^{\rm{2}}}}}\;{\rm{ = }}\,{\rm{3}}{\rm{.74  \times  1}}{{\rm{0}}^{\rm{5}}}\]

Taking the square root of both sides we obtain:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}}{{{\rm{0}}{\rm{.5x}}}}\;{\rm{ = }}\,{\rm{612}}\]

x = 0.00136

Therefore, the equilibrium concentrations are:

[H₂] = [I₂] = 0.5 x 0.00136 = 0.00068 = 6.8 x 10^-4 mol/L

[HI] = 0.417 – 0.00136 = 0.416 mol/L

Check:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{0.416}^2}}}{{{{0.00068}^2}}}\; = \,3.74{\rm{ }} \times {\rm{ }}{10^5}\]

✔

The equilibrium constant for this reaction is:

\[{K_p}\; = \;\frac{{{P_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}}}{{{P^2}_{{{\rm{H}}_{\rm{2}}}}{P_{{\rm{CO}}}}}}\; = \;6.7{\rm{ }} \times {\rm{ }}{10^{ – 3}}\]

Now, if the mixture is at equilibrium, then plugging the concentration values in this equation, we should get the value for Kp.

\[{K_p}\; = \;\frac{{0.598}}{{{{0.248}^2}0.085}}\; = \;114\; > \;6.7{\rm{ }}x{\rm{ }}{10^{ – 3}}\]

This is greater than Kp (6.7 x 10^-3), and therefore, the reaction will proceed to the left forming more CO and H_2. This can also be predicted when we compare the partial pressures of the gases. Despite a small equilibrium constant, CH₃OH has a higher partial pressure than CO and H_2. The system, cannot stay in this state, and thus some of the CH₃OH will break into CO and H_2.

To set up the ICE table, we need to determine which gas will increase and which one will decrease. Because only N₂O₄ is initially present in the system, then its concentration is going to decrease, while the concentration of NO₂ will increase.

Let’s suppose x mol/L of N₂O₄ decomposes (reacts) to form 2x mol/L NO_2. This is based on the stoichiometric ratio of the gases – every one mole of N₂O₄ produces 2 mol NO₂.

N₂O₄(g) ⇆ 2NO₂(g)

The equilibrium constant for this reaction is:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left[ {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right]}^2}}}{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_4}} \right]}}\; = \,0.654\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{{\rm{(2x)}}}^2}}}{{0.0600 – x}}\; = \,0.654\]

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{{\rm{4x}}}^2}}}{{0.0600 – x}}\; = \,0.654\]

4x² + 0.654x – 0.003924 = 0

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]

a = 4, b = 0.654, c = -0.003924

x=0.0467 or x=−0.210 (doesn’t work)

Therefore, the equilibrium concentrations are:

[N₂O₄] = 0.0600 – 0.0467 = 0.0133 mol/L

[NO₂] = 2 x 0.0467 = 0.0934 mol/L

The equilibrium constant for this reaction is:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left[ {{\rm{PC}}{{\rm{l}}_{\rm{5}}}} \right]}}{{\left[ {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}\;{\rm{ = }}\,{\rm{255}}\]

To set up the ICE table, we need to determine which gas will increase and which one will decrease. Because no PCl₅ is present in the system, the reaction will proceed to the right, thus decreasing the concentration of PCl₃ and Cl_2.

Let’s suppose x mol/L of PCl₃ and Cl₂ reacts to form x mol/L PCl_5. This is based on the stoichiometric ratio of the gases – every one mole of PCl₃ and Cl₂ produces one mol PCl₅.

PCl₃(g) + Cl₂(g) ⇆ PCl₅(g)

Plugging the numbers, we get an equation based on the equilibrium constant:

\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\rm{x}}}{{\left( {{\rm{0}}{\rm{.3500 – x}}} \right)\left( {{\rm{0}}{\rm{.375 – x}}} \right)}}\;{\rm{ = }}\,{\rm{255}}\]

255x² – 185.875x + 33.46875 = 0

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]

a = -255, b = 185.9, c = -33.47

x=0.324691 or x=0.404231

x=0.404231 doesn’t work becasue 0.3500-0.404231 gives a negative value for the concentration.

x=0.3247

Therefore, the equilibrium concentrations are:

[PCl₃] = 0.3500 – 0.3247 = 0.0253 mol/L

[Cl₂] = 0.375 – 0.3247 = 0.0503 mol/L

[PCl₅] = x=0.3247  mol/L

Remember, when converting Kc and Kp, the value for R is most often used as 0.08206 L atm/ mol K, and therefore, the units for pressure should also be in atm.

\[{P_{({{\rm{I}}_{\rm{2}}})}}\; = \;{\rm{655}}\;{\rm{torr\;}} \times {\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}{\rm{\; = }}\,{\rm{0}}{\rm{.862}}\;{\rm{atm}}\]

\[{P_{({\rm{C}}{{\rm{l}}_{\rm{2}}})}}\; = \;{\rm{864}}\;{\rm{torr\;}} \times {\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}{\rm{\; = }}\,{\rm{1}}{\rm{.14}}\;{\rm{atm}}\]

The equilibrium constant is:

\[{K_p}\; = \;\frac{{{P^2}_{{\rm{ICl}}}}}{{{P_{{{\rm{I}}_{\rm{2}}}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\; = \;2.34{\rm{ }} \times {\rm{ }}{10^{ – 4}}\]

However, remember that these are the partial pressures at equilibrium so, we cannot use the numbers above.

First, we need to set up an ICE table to determine the partial pressures at equilibrium. Because the system does not contain any ICl at the beginning, its pressure (concentration) is going to increase, while the pressure of Cl₂ and I₂ is going to decrease. Let’s assume the pressure of Cl₂ and I₂ drops by x atm which means the pressure of ICl will increase by 2x atm:

Plug in the numbers and assign x for the partial pressure of ICl:

\[{K_p}\; = \;\frac{{{{{\rm{(2x)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862 – x}}} \right)\left( {{\rm{1}}{\rm{.14 – x}}} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

\[{K_p}\; = \;\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}\;{\rm{ – }}\;{\rm{2x}}\;{\rm{ + }}\;{\rm{098268}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Solving this quadratic equation, we get x=0.00753

Therefore, the partial pressure of ICl at equilibrium is 2 x 0.00753 atm = 0.0141 atm = 11.4 torr

A good approach to avoid solving long quadratic equations is to make an approximation/assumption that x << 0.862, meaning that there is very little pressure drop (reaction).

In this case, we can simplify the quadratic equation as:

\[{K_p}\; = \;\frac{{{{{\rm{(2x)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862}}} \right)\left( {{\rm{1}}{\rm{.14}}} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

Solving for x, we find that x = 0.007582 which is very close to 0.00753 (error < 1%).

Let’s assume though that we did not know the actual value for x and only solve for it by the approximation. To test whether the assumption was applicable to this reaction, plug the values in the equilibrium expression and compare the result to the actual Kc given in the problem.

\[{K_p}\; = \;\frac{{{{{\rm{(2}} \times 0.007582{\rm{)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862 – }}0.007582} \right)\left( {{\rm{1}}{\rm{.14 – }}0.007582} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.38  \times  1}}{{\rm{0}}^{{\rm{ – 4}}}}\]

And this confirms that the approximation was suitable for this problem. If it wasn’t, and the difference between the Kc values was more than 5%, then the equation must be solved without the approximation.

Since the system initially contains both reactants and products, we need to first determine the reaction quotient:

\[Q\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;\]

\[Q\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.450 }} \times \;{\rm{0}}{\rm{.250}}}}{{0.650}}\; = 0.173\]

Q < K, so, the reaction will proceed to the right. This means, the concentrations of POCl and Cl₂ are going to increase, while the concentration of POCl₃ will decrease.

Let’s set up an ICE table and assign x as the depletion in the concentration of POCl₃:

POCl3(g) ⇌ POCl(g) + Cl₂(g)

The equilibrium constant is:

\[K\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;\]

\[K\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.450 + x}}} \right)\left( {{\rm{0}}{\rm{.250 + x}}} \right)}}{{0.650 – x}}\; = 0.650\]

x=0.200

Therefore, the [POCl] at equilibrium is 0.450 + 0.200 = 0.650 mol/L

The reaction is going to proceed to the right because there are only reactants present in the mixture initially (Q = 0).

Even though we don’t need an ICE table here, let’s put one to visualize the changes. Here is the date we have:

2NO(g) + 2H₂(g) ⇆ N₂(g) + 2H₂O(g)

Next is to determine how many moles of NO have reacted to establish the equilibrium. For this, subtract the final (equilibrium) and initial concentrations of NO:

Change (NO) = 0.15-0.056=0.094 mol

Add this to the ICE table and determine the moles of NO based on the stoichiometric ratio. Every 2 moles of NO produce one mole of N₂, and therefore,

n (N2) = 0.094/2 = 0.047 mol

2NO(g) + 2H₂(g) ⇆ N₂(g) + 2H₂O(g)

And since there is no N₂ present in the beginning, this is also the number of moles that are present at equilibrium.

The first here is to determine the direction of the reaction. Now, there are both reactants and products, however, there is no NO, and remember, when one of the reactants or products is missing from the system, the equilibrium is going to be established by producing some of that compound. Therefore, in this case, the reaction is going to proceed to the right to produce some NO.

So, let’s set up an ICE table assigning 2x as the depletion in NOCl concentration:

 2NOCl(g) ⇆ 2NO(g) + Cl₂(g)

The equilibrium constant is:

\[K = \frac{{{{[{\rm{NO}}]}^{\rm{2}}}[{\rm{C}}{{\rm{l}}_{\rm{2}}}]}}{{{{[{\rm{NOCl}}]}^{\rm{2}}}}}\]

\[K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + x}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2x}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{–6}}}}\]

This is a cubic equiation, and we can approach it by an approximation that 2x<<2.00 or x<<1.00. This allows to simplify the equation as follows:

\[K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{–6}}}}\]

\[K = \;\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65}}} \right)}}{{\rm{4}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{–6}}}}\]

6.6x² = 9.6×10^-6

x=0.00121

The equilibrium concentratio of NO is 2x, so that is 0.00242 mol/L.

Now, we need to plug the numbers and see if the value for the equilibrium constant is close enough to the given number (K = 2.4 x 10^–6).

\[K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + x}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2x}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4  \times  1}}{{\rm{0}}^{{\rm{–6}}}}\]

\[K = \;\frac{{{{\left( {{\rm{2}} \times {\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + 0}}{\rm{.00121}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2}} \times {\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.42  \times  1}}{{\rm{0}}^{{\rm{–6}}}}\]

And that is very close to the value of K, and therefore, out approximation was correct and the equilibrium concentratio of NO is 0.00242 mol/L.

(a) the temperature is increased: We need to first realize that this is an endothermic reaction (ΔH^o positive). This means the reaction consumes heat, or, in other words, heat is a reactant. According to Le Châtelier’s principle, if we increase the temperature, the reaction will go in the direction that absorbs heat. Therefore, the reaction will shift right.

(b) the pressure on the gases is decreased: The number of gas molecules is different, so the reaction will respond in a way that brings the pressure up again. The products contain more gas molecules, therefore, the reaction will shift to the right.

(c) the concentration of O₂ is increased: The reaction will shift left to use up some of the added oxygen gas and decrease its concentration.

(d) a is added catalyst: The catalyst is used to speed up the reaction so it reaches equilibrium more quickly. However, it has no effect on the equilibrium position.

(e) an inert gas is added at constant volume: Remember, changing the pressure of a gaseous reaction only has an effect on the equilibrium position if it results in changes of partial pressures of gases in the chemical equation. And although adding an inert gas to the system, increases the total pressure, the partial pressures of gases do not change. Therefore, it has no effect on the equilibrium composition.

a) decreasing the reaction volume – Has no effect since the number of gas molecules is equal on both sides.

b) removing C₃H₆Cl₂ from the reaction mixture as it forms – Will increase the yield of the reaction since the equilibrium will shift to the right to reinstate the amount of product removed from the system.

c) adding a catalyst – Has no effect on the equilibrium position

d) adding Cl₂ – Will shift the reaction right and thus, increase the amount of product.

If we increase the volume of a reaction container, the reaction itself will shift to the side that contains more gas molecules since this increases its own volume. If there are an equal number of gas molecules on both sides of the reaction, then the reaction will remain at equilibrium regardless of how we change the volume of the container.

1) The reaction will shift left producing more reactants because there are 2 molecules of gas reactants compared with 1 molecule of gas on the product side.

2) The reaction will shift right producing more products because there are 4 molecules of gas products compared with 2 molecules of gas on the reactants side.

3) The equilibrium will not change because both sides contain an equal number of molecules in gas state.

4) The equilibrium will not change because both sides contain an equal number of molecules in gas state.

5) Only the products contain a gas molecule, therefore, the reaction will shift right. Remember, we ignore the solids and liquids and only concentrate on the gases because of their extremely large volume compared with solids and liquids.

(a) the temperature is raised: We need to first realize that this is an endothermic reaction (ΔH^o positive). This means the reaction consumes heat, or, in other words, heat is a reactant. According to Le Châtelier’s principle, if we increase the temperature, the reaction will go in the direction that absorbs heat. Therefore, the reaction will shift right.

(b) more CO gas is added to the reaction mixture: The reaction will shift right to consume some of the added CO gas and decrease its concentration.

(c) some CO₂ is removed from the mixture: The reaction will shift right to reinstate the original concentration of CO₂.

(d) the pressure on the gases is increased: If the number of gas molecules was different, the reaction would respond in a way that brings the pressure back down. However, the number of gas molecules is equal here, therefore, there will be no effect on the equilibrium.

(e) a catalyst is added to the reaction mixture: The catalyst is used to speed up the reaction so it reaches equilibrium more quickly. However, it has no effect on the equilibrium position.

According to the Le Châtelier’s principle, when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the system shifts in the direction that counteracts this change.

a) NO is added: The reaction will shift left to use up some of the added nitrogen monoxide and decrease its concentration.

b) N₂ is added: The reaction will shift right to consume some of the added nitrogen gas and decrease its concentration.

c) NO is removed: The reaction will shift right to reinstate the original concentration of NO.