Two solids dissociate as follows:A_{(s)} righ | vedclass

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(A) Let the partial pressure of $B_{(g)}$ be $P_1$ and $E_{(g)}$ be $P_2$.From the first equilibrium: $A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$,t

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$2 \sqrt {x + y} \, atm$

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(A) Let the partial pressure of $B_{(g)}$ be $P_1$ and $E_{(g)}$ be $P_2$.From the first equilibrium: $A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$,the partial pressure of $C_{(g)}$ contributed by this reaction is $P_1$.From the second equilibrium: $D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$,the partial pressure of $C_{(g)}$ contributed by this reaction is $P_2$.Total partial pressure of $C_{(g)} = P_1 + P_2$.$K_{p_1} = P_B \cdot P_C = P_1(P_1 + P_2) = x$$K_{p_2} = P_C \cdot P_E = (P_1 + P_2)P_2 = y$Adding both equations:$x + y = P_1(P_1 + P_2) + P_2(P_1 + P_2) = (P_1 + P_2)(P_1 + P_2) = (P_1 + P_2)^2$So,$(P_1 + P_2) = \sqrt {x + y}$.Total pressure $P_{total} = P_B + P_C + P_E = P_1 + (P_1 + P_2) + P_2 = 2(P_1 + P_2)$.Substituting the value: $P_{total} = 2 \sqrt {x + y} \, atm$.

Two solids dissociate as follows:
$A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$; $K_{p_1} = x \, atm^2$
$D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$; $K_{p_2} = y \, atm^2$
The total pressure when both the solids dissociate simultaneously is:

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Two solids dissociate as follows:$A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$; $K_{p_1} = x \, atm^2$$D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$; $K_{p_2} = y \, atm^2$The total pressure when both the solids dissociate simultaneously is: