In the reaction,A_{(s)} + B_{(g)} rightleftha | vedclass

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(C) For the reaction $A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,the equilibrium constant $K_p$ is given by $K_p = \frac{P_D^2}{P_B}$.S

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$\sqrt{2}$

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(C) For the reaction $A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,the equilibrium constant $K_p$ is given by $K_p = \frac{P_D^2}{P_B}$.Since $K_p$ is constant at a constant temperature,we have $\frac{P_{D,1}^2}{P_{B,1}} = \frac{P_{D,2}^2}{P_{B,2}}$.Given that the pressure of $B$ is doubled,$P_{B,2} = 2P_{B,1}$.Substituting this into the equation: $\frac{P_{D,1}^2}{P_{B,1}} = \frac{P_{D,2}^2}{2P_{B,1}}$.This simplifies to $P_{D,2}^2 = 2P_{D,1}^2$,which means $P_{D,2} = \sqrt{2} P_{D,1}$.Therefore,the pressure of $D$ becomes $\sqrt{2}$ times the initial pressure.

In the reaction,$A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,equilibrium is established. If the pressure of $B$ is doubled,then to reestablish the equilibrium,how many times does the pressure of $D$ become as compared to its initial pressure?

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