Question 71

Dissociation of a gas $$A_{2}$$ takes place according to the following chemical reaction.
At equilibrium, the total pressure is 1 bar at 300K.

$$A_{2}(g)\rightleftharpoons 2A(g)$$

The standard Gibbs energy of formation of the involved substances has been
provided below:

The degree of dissociation of $$A_{2} (g)$$ is given by $$(x\times10^{-2})^{1/2} $$ where $$x$$ =
_____ . (Nearest integer).
[Given: $$R=8 J \text{ }mol^{-1}K^{-1},\log{2}=0.3010, \log {3}=0.48]$$
Assume degree of dissociation is not negligible.

Correct Answer: 33

Solution

Find $$\Delta G^\circ_r$$: We use the values provided in the table ($$2 \times \text{Products} - \text{Reactants}$$). Note that $$\Delta G^\circ_r = -1.664 \text{ kJ/mol}$$.

Find $$K_p$$: Using the relation $$\Delta G^\circ = -RT \ln K_p$$. Plugging in $$R=8$$, $$T=300$$, and converting kJ to J, we find $$\ln K_p = 0.693$$. Since $$\ln 2 \approx 0.693$$, then $$K_p = 2$$.

Relate $$K_p$$ to $$\alpha$$: For the dissociation $$A_2 \rightleftharpoons 2A$$, the partial pressures are derived from the total pressure ($$P=1$$). The formula simplifies to:

$$K_p = \frac{4\alpha^2}{1-\alpha^2} \times P$$

Solve for $$x$$: Substituting $$K_p = 2$$ and $$P = 1$$, we get $$2 - 2\alpha^2 = 4\alpha^2$$, which leads to $$\alpha = \sqrt{1/3}$$.

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