x A_{(s)} rightleftharpoons y B_{(g)} + z C_{ | vedclass

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(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c 	imes (RT)^{\Delta n_g}$.Rearranging this,we get $\frac{K_p}{K_c} = (RT)^{\Delta

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$y + z = 2$

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(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c 	imes (RT)^{\Delta n_g}$.Rearranging this,we get $\frac{K_p}{K_c} = (RT)^{\Delta n_g}$.Given that $\frac{K_c}{K_p} = (RT)^{-2}$,we can invert this to find $\frac{K_p}{K_c} = (RT)^{2}$.Comparing the exponents,we find $\Delta n_g = 2$.For the reaction $x A_{(s)} ightleftharpoons y B_{(g)} + z C_{(g)}$,the change in gaseous moles is $\Delta n_g = (y + z) - 0 = y + z$.Therefore,$y + z = 2$.

$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?

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