Equilibria: Question 7

Worked solution

Part (a): Writing the Kp expression

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

Part (b): Equilibrium amounts and mole fractions

$60.0\%$ of the initial $0.0500\ \text{mol}$ of $\text{N}_2\text{O}_4$ has dissociated: $n(\text{N}_2\text{O}_4\text{ dissociated}) = 0.600\times0.0500 = 0.0300\ \text{mol}$

So the amount of $\text{N}_2\text{O}_4$ remaining at equilibrium is: $n(\text{N}_2\text{O}_4) = 0.0500 - 0.0300 = 0.0200\ \text{mol}$

Since the stoichiometry is $1:2$, each mole of $\text{N}_2\text{O}_4$ that dissociates forms 2 moles of $\text{NO}_2$: $n(\text{NO}_2) = 2\times0.0300 = 0.0600\ \text{mol}$

The total number of moles of gas at equilibrium is: $n_{total} = 0.0200 + 0.0600 = 0.0800\ \text{mol}$

(Check: this is greater than the initial 0.0500 mol, as expected, since dissociation increases the total number of gas particles.)

The mole fractions are: $x(\text{N}_2\text{O}_4) = \frac{0.0200}{0.0800} = 0.250 \qquad x(\text{NO}_2) = \frac{0.0600}{0.0800} = 0.750$

(Check: $0.250+0.750=1.00$. Consistent.)

Part (c): Partial pressures

Each partial pressure is the mole fraction multiplied by the total equilibrium pressure, $100\ \text{kPa}$: $p(\text{N}_2\text{O}_4) = 0.250\times100 = 25\ \text{kPa}$ $p(\text{NO}_2) = 0.750\times100 = 75\ \text{kPa}$

(Check: $25+75=100\ \text{kPa}$, matching the given total pressure.)

Part (d): Calculating Kp

Substituting the partial pressures from part (c) into the expression from part (a): $K_p=\frac{(75)^2}{25}=\frac{5625}{25}=225$

(Check: recomputing independently, $75\times75=5625$, and $5625\div25=225$. Consistent.)

Units: $\text{NO}_2$ has an exponent of 2, giving $\text{kPa}^2$ in the numerator; the denominator is $\text{kPa}^1$. These do not cancel fully, leaving a net power of $\text{kPa}^{2-1}=\text{kPa}^1$. So: $K_p = 225\ \text{kPa}$

Final answers

• (a) $K_p=\dfrac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)}$
• (b) $n(\text{N}_2\text{O}_4)=0.0200\ \text{mol}$, $n(\text{NO}_2)=0.0600\ \text{mol}$; $x(\text{N}_2\text{O}_4)=0.250$, $x(\text{NO}_2)=0.750$
• (c) $p(\text{N}_2\text{O}_4)=25\ \text{kPa}$, $p(\text{NO}_2)=75\ \text{kPa}$
• (d) $K_p=225\ \text{kPa}$