Question

Thermal decomposition of gaseous $X_{2}$ to gaseous $X$ at $298 K$ takes place according to the following equation:

$X_{2(g)} \rightleftharpoons 2 X_{(g)}$

The standard reaction Gibbs energy, $\Delta_{r} G^{\circ},$ of this reaction is positive. At the start of the reaction, there is one mole of $X_{2}$ and no $X$. As the reaction proceeds, the number of moles of $X$ formed is given by $\beta$. Thus, $\beta_{\text {equilibrium }}$ is the number of moles of $X$ formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.

(Given $: R=0.083 L$ bar $\left. K ^{-1} mol ^{-1}\right)$

The incorrect statement among the following, for this reaction, is

• A. decrease in the total pressure will result in formation of more moles of gaseous $X$
• B. at the start of the reaction, dissociation of gaseous $X_{2}$ takes place spontaneously
• C. $\beta_{\text {equlibrium }}=0.7$
• D. $K_{c}<1$

Answer 1

Answer: (C)

(a) If the pressure on the system is decreased, the equilibrium will shift in the direction in which pressure increases i.e., increase in no. of moles takes place i.e., in forward direction.

(b) At the start of the reaction, $Q < K$ thus, the reaction will proceed in the forward direction i.e., reaction is spontaneous.

(c) If $\beta_{e q}=0.7$ then, $K_{p}=\frac{8 \times(0.7)^{2}}{4-(0.7)^{2}}>1$

$\Delta G^{\circ}=-R T \ln K_{p}$ so, $\Delta G^{\circ}=-$ ve but given $\Delta G^{\circ}=+$ ve so, $K_{p}$

should be less than 1 hence, $\beta_{e q} \neq 0.7$.

(d) $K_{p}=K_{c}(R T)^{\Delta n}$

$K_{c} < K_{p} \quad(\because R T>1)$

If $K_{p} < 1$ then $K_{c}<1$