Question

Question: The gaseous compounds allene and propyne are isomers with formula C₃H₄. Calculate the equilibrium constant and the standard enthalpy change at 25°C for the isomerization reaction Allene(g)$\rightleftharpoons$Propyne(g)from the following data, all of which apply to 298 K:

	$△{H_{\mathrm{f}}}^{°}(kJmo{l}^{-1})$	$△{G_f}^{°}(kJmo{l}^{-1})$
Allene	192	202
Propyne	185	194

Short answer

The equilibrium constant for the isomerization reaction is 24.53 and the standard enthalpy change for the given reaction is -7 KJ/mol.

Step-by-step solution

Step-by-step solution

Step 1:

By using the Gibbs free energy change we can calculate the equilibrium constant as,

$$\begin{gathered} ∆{\mathrm{G}}^{°}=∆{{{\mathrm{G}}^{}}_{\mathrm{f}}}^{°}\left(\mathrm{Product}\right)-∆{{\mathrm{G}}_{\mathrm{f}}}^{°}\left(\mathrm{Reactant}\right) \\ =194-202 \\ =-8\mathrm{KJ}/\mathrm{mol} \end{gathered}$$

Now, substituting the vale in the following reaction,

$$\begin{gathered} \mathrm{\Delta G}°=-\mathrm{RTlnK} \\ -8\mathrm{KJ}/\mathrm{mol}=-\left(8.314\times {10}^{-3}\mathrm{KJ}/\mathrm{mol}.\mathrm{K}\right)\left(298\mathrm{K}\right)\mathrm{lnk} \\ \mathrm{lnk}=\frac{8\mathrm{KJ}/\mathrm{mol}}{\left(8.314\times {10}^{-3}\mathrm{KJ}/\mathrm{mol}.\mathrm{K}\right)\left(298\mathrm{K}\right)} \\ =0.0032\times {10}^3 \\ =3.2 \\ \mathrm{k}={\mathrm{e}}^{3.2}=24.53 \end{gathered}$$

Hence, the equilibrium constant for the isomerization reaction is 24.53.

Step 2:

The standard enthalpy change for the given reaction is as follows,

$$\begin{gathered} ∆{\mathrm{H}}^{°}=∆{{\mathrm{H}}_{\mathrm{f}}}^{°}\left(\mathrm{Product}\right)-∆{{\mathrm{H}}_{\mathrm{f}}}^{°}\left(\mathrm{Reactant}\right) \\ =185-192 \\ =-7\mathrm{KJ}/\mathrm{mol} \end{gathered}$$

Hence, the standard enthalpy change for the given reaction is -7 KJ/mol.