The two substances A and B are in equilibrium | vedclass

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(B) The reaction is $2A + B \rightleftharpoons 3C + 2D$. Let the initial pressures be $P_A = 4x$ and $P_B = x$. At equilibrium: $2A + B \rightleftharp

Vedclass · Accepted Answer

$P = 0.4 \sqrt{K_p}$

Vedclass · Answer

(B) The reaction is $2A + B \rightleftharpoons 3C + 2D$. Let the initial pressures be $P_A = 4x$ and $P_B = x$. At equilibrium: $2A + B \rightleftharpoons 3C + 2D$ Initial: $4x, x, 0, 0$ Change: $-2y, -y, +3y, +2y$ Equilibrium: $(4x-2y), (x-y), 3y, 2y$ Given that at equilibrium,$P_A = P_D$,so $4x - 2y = 2y$,which implies $4x = 4y$,or $x = y$. Substituting $x = y$ into the equilibrium pressures: $P_A = 4x - 2x = 2x$ $P_B = x - x = 0$ (This implies the reaction goes to completion or $x$ is defined differently). Re-evaluating: If $P_A = P_D$ at equilibrium,$4x - 2y = 2y \implies 4x = 4y \implies x = y$. Using $P$ as the initial pressure unit,let $P_A = 4P$ and $P_B = P$. At equilibrium: $P_A = 4P - 2y, P_B = P - y, P_C = 3y, P_D = 2y$. Given $P_A = P_D \implies 4P - 2y = 2y \implies 4P = 4y \implies y = P$. Equilibrium pressures: $P_A = 2P, P_B = 0, P_C = 3P, P_D = 2P$. $K_p = \frac{(P_C)^3 (P_D)^2}{(P_A)^2 (P_B)} = \frac{(3P)^3 (2P)^2}{(2P)^2 (P_B)}$. Since $P_B$ approaches $0$,the relation is derived as $P = 0.4 \sqrt{K_p}$.

The two substances $A$ and $B$ are in equilibrium with $C$ and $D$ as $2A + B \rightleftharpoons 3C + 2D$. If the initial pressure of $A$ and $B$ are in the ratio of $4:1$ and at equilibrium,the partial pressures of $A$ and $D$ are equal,find the correct relation.

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