Given K_p for the reaction frac{1}{2} C_{(g)} | vedclass

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(A) For the reaction,$\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$,given $K_{p1} = 0.25$. For the target reaction

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) For the reaction,$\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$,given $K_{p1} = 0.25$. For the target reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$,let the equilibrium constant be $K_{p2}$. First,reverse the given reaction: $\frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)} \rightleftharpoons \frac{1}{2} C_{(g)}$. The new constant $K'_{p} = \frac{1}{K_{p1}} = \frac{1}{0.25} = 4$. Next,multiply the stoichiometry by $2$ to match the target reaction: $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$. The constant becomes $K_{p2} = (K'_{p})^2 = (4)^2 = 16$. Thus,$K_p$ for the reaction is $16$.

Given $K_p$ for the reaction $\frac{1}{2} C_{(g)} \rightleftharpoons \frac{1}{2} A_{(g)} + \frac{1}{2} B_{(g)}$ at a fixed temperature is $0.25 \ atm$. Find the $K_p$ for the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$ at the same temperature.

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