Question

Thermal decomposition of gaseous $X_2$ to gaseous X at 298 K takes place according to the following equation:
$X_2\left(g\right)\rightleftharpoons 2X\left(g\right)$
The standard reaction Gibbs energy, ${\Delta }_r{G}^0,$ 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, ${\beta }_{equilibrium}$ 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 $K^{-1}mo{l}^{-1}$)

The equilibrium constant $K_p$ for this reaction at 298 K, in terms of ${\beta }_{equilibrium},$ is:

• A. $\frac{8{\beta }_{equilibrium}^2}{2-{\beta }_{equilibrium}}$
• B. $\frac{8{\beta }_{equilibrium}^2}{4-{\beta }_{equilibrium}^2}$
• C. $\frac{4{\beta }_{equilibrium}^2}{2-{\beta }_{equilibrium}}$
• D. $\frac{4{\beta }_{equilibrium}^2}{4-{\beta }_{equilibrium}^2}$

Answer 1

The correct option is B $\frac{8{\beta }_{equilibrium}^2}{4-{\beta }_{equilibrium}^2}$

$X_2\left(g\right)\rightleftharpoons 2X\left(g\right)101-\frac{{\beta }_e}{2}{\beta }_e$
Total number of moles at equilibrium
$\Rightarrow 1-\frac{{\beta }_e}{2}+{\beta }_e$
$\Rightarrow 1+\frac{{\beta }_e}{2}$
$K_p=\frac{(p_x{)}^2}{p_{x_2}}$
$=\frac{\frac{(\frac{{\beta }_e\times 2}{1+\frac{{\beta }_e}{2}}{)}^2}{(1-\frac{{\beta }_e}{2})\times 2}}{1+\frac{{\beta }_e}{2}}=\frac{\frac{2{\beta }_e^2}{1-\frac{{\beta }_e^2}{4}}}{4}{K}_p=\frac{8{\beta }_e^2}{4-{\beta }_e^2}$