Chemical equilibria

WCH14Topic 4.33 questions

Equilibrium constants $K_c$ and $K_p$ and the effect of conditions on equilibria.

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Question 14 marks
At a certain temperature, an equilibrium mixture of $\text{N}_2\text{O}_4$ and $\text{NO}_2$ has a total pressure of $2.00\,\text{atm}$ . The mole fraction of $\text{NO}_2$ is $0.400$ .

$\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})$

Calculate $K_p$ , giving its units.
Mark scheme
- Partial pressure = mole fraction $\times$ total pressure (1)
- $p(\text{NO}_2) = 0.800\,\text{atm}$ and $p (N_{2} O_{4}) = 0.600 \times 2.00$ (1)
Question 25 marks
A mixture of $0.500\,\text{mol}$ ethanoic acid and $0.800\,\text{mol}$ ethanol was left with a few drops of acid catalyst until equilibrium was reached. The equilibrium mixture contained $0.380\,\text{mol}$ of ethyl ethanoate.

${CH}_{3} COOH + C_{2} H_{5} OH ⇌ {CH}_{3} {COOC}_{2}$

(

)

=

1.20

atm

p(\text{N}_2\text{O}_4) = 0.600 \times 2.00 = 1.20\,\text{atm}

p (N_{2} O_{4}) = 0.600 \times 2.00 = 1.20 atm

K_p = \dfrac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)}

(1)

K_p = \dfrac{0.800^2}{1.20} = 0.533\,\text{atm}

(1)

H_{5}

+

H_{2}

O

\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}

CH_{3} COOH + C_{2} H_{5} OH ⇌ CH_{3} COOC_{2} H_{5} + H_{2} O

Calculate $K_c$ for this reaction. Ignore any water added with the catalyst.

Mark scheme

Equilibrium amounts: ethanoic acid $0.120\,\text{mol}$ and ethanol $0.420\,\text{mol}$ (1)
Equilibrium amount of water $= 0.380\,\text{mol}$ (1)
$K_c = \dfrac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$ (1)
The volume cancels because there are equal numbers of moles on each side, so amounts can be used directly (1)
$K_c = \dfrac{0.380 \times 0.380}{0.120 \times 0.420} = 2.87$ (no units) (1)

]

[CH3COOC2H5][H2O]

2.87