Question

If $K_1$ and $K_2$ are the respective equilibrium constants for the two reactions. ${XeF6_{(g)} +H2O_{(g)}<=> XeOF4_{(g)} +2HF_{(g)}}$ ${XeO4_{(g)} +XeF6_{(g)} <=> XeOF4_{(g)} +XeO3F2_{(g)}}$ The equilibrium constant for the reaction, ${XeO4_{(g)} +2HF_{(g)} <=> XeO3F2_{(g)} +H2O_{(g)}}$ is

• A. $K_1K_2$
• B. $K_1/K^2_2$
• C. $K_2/K_1$
• D. $K_1/K^2_2$

Answer 1

Answer: (C)

$K_2/K_1$

Answer 2

$K_{1}=\frac{\left[XeOF_{4}\right]\left[HF\right]^{2}}{\left[XeF_{6}\right]\left[H_{2}O\right]} \ldots\left(i\right)$ $K_{2}=\frac{\left[XeOF_{4}\right]\left[XeO_{3}F_{2}\right]}{\left[XeO_{4}\right]\left[XeF_{6}\right]}\ldots\left(ii\right)$ Dividing E $\left(ii\right)$ by $\left(i\right)$ we have $\frac{K_{2}}{K_{1}}=\frac{\left[XeO_{3}F_{2}\right]\left[H_{2}O\right]}{\left[XeO_{4}\right]\left[HF\right]^{2}}=K'$