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, $\Delta G^o,$ 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

$\frac{8{\beta }_{equilibrium}^2}{4-{\beta }_{equilibrium}^2}$

$X_2\to 2X\left(g\right)$ $△G^o>0$
1 0
$1-{\beta }_{eq/2}$ ${\beta }_{eq}$ $-RTIn{K}_{eq}>0$
$RTIn{K}_{eq}<0$
$2.303RTlog\left(\frac{{\beta }^2}{1-{\beta }^2/2^2}\right).2<0$
$k_p=\frac{2{\beta }^2}{1-{\beta }^2/4}=\frac{8{\beta }^2}{4-{\beta }^{equilibrium}}$