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 $K_p = K_c(RT)^{\Delta n_g}$.Given the expression $\frac{K_c}{K_p} = (RT)^{-2}$,we can rewrit

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$y + z = 2$

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(D) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.Given the expression $\frac{K_c}{K_p} = (RT)^{-2}$,we can rewrite this as $\frac{K_p}{K_c} = (RT)^2$.Comparing this with $K_p = K_c(RT)^{\Delta n_g}$,we get $\Delta n_g = 2$.For the reaction $x A_{(s)} ightleftharpoons y B_{(g)} + z C_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = (y + z) - (	ext{moles of gaseous reactants})$.Since $A$ is a solid,its moles are not included in $\Delta n_g$. Thus,$\Delta n_g = y + z$.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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