1. Tardigrade
  2. Question
  3. Chemistry
  4. Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation: X2(g) leftharpoons 2 X(g) The standard reaction Gibbs energy, Δr G°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, β 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 . K -1 mol -1) The equilibrium constant Kp for this reaction at 298 K , in terms of β text equilibrium , is

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Q. 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 equilibrium constant $K_{p}$ for this reaction at $298 K ,$ in terms of $\beta_{\text {equilibrium }},$ is

Equilibrium

Solution:

image

$K_{p}=\frac{\left(p_{X}\right)^{2}}{\left(p_{X_{2}}\right)} ; p_{X}=\left(\frac{\beta_{e q}}{1-\frac{\beta_{e q}}{2}+\beta_{e q}}\right) P_{\text {total }}=\left(\frac{\beta_{e q}}{1+\frac{\beta_{e q}}{2}}\right) P_{\text {total }}$

$p_{X_{2}}=\left(\frac{1-\frac{\beta_{e q}}{2}}{1+\frac{\beta_{e q}}{2}}\right) P_{\text {total }}$

$K_{p}=\frac{\left[\left(\frac{\beta_{e q}}{1+\frac{\beta_{e q}}{2}}\right) P_{\text {total }}\right]^{2}}{\left(\frac{1-\frac{\beta_{e q}}{2}}{1+\frac{\beta_{e q}}{2}}\right) P_{\text {total }}}=\left(\frac{\beta_{e q}^{2}}{1-\frac{\beta_{e q}^{2}}{4}}\right) P_{\text {total }}$

$=\left(\frac{4 \beta_{e q}^{2}}{4-\beta_{e q}^{2}}\right) P_{\text {total }}=\left(\frac{4 \beta_{e q}^{2}}{4-\beta_{e q}^{2}}\right) \times 2=\frac{8 \beta_{e q}^{2}}{4-\beta_{e q}^{2}}$