For the reaction 2AB_{(g)} rightleftharpoons | vedclass

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(B) The reaction is $2AB_{(g)} \rightleftharpoons 2A_{(g)} + B_{2(g)}$.Initial pressures: $P_{AB} = 100 \ atm$,$P_A = 0$,$P_{B_2} = 0$.Let the decreas

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) The reaction is $2AB_{(g)} ightleftharpoons 2A_{(g)} + B_{2(g)}$.Initial pressures: $P_{AB} = 100 \ atm$,$P_A = 0$,$P_{B_2} = 0$.Let the decrease in pressure of $AB$ be $2x$.At equilibrium: $P_{AB} = 100 - 2x$,$P_A = 2x$,$P_{B_2} = x$.Total pressure at equilibrium: $(100 - 2x) + 2x + x = 125 \ atm$.$100 + x = 125 \implies x = 25 \ atm$.Equilibrium partial pressures: $P_{AB} = 100 - 2(25) = 50 \ atm$,$P_A = 2(25) = 50 \ atm$,$P_{B_2} = 25 \ atm$.$K_p = \frac{(P_A)^2 	imes (P_{B_2})}{(P_{AB})^2} = \frac{(50)^2 	imes 25}{(50)^2} = 25$.

For the reaction $2AB_{(g)} \rightleftharpoons 2A_{(g)} + B_{2(g)}$; if the initial pressure of $AB$ is $100 \ atm$ and at equilibrium the total pressure becomes $125 \ atm$,the equilibrium constant $(K_p)$ will be:

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