For a reaction,N_2O_{5(g)} rightarrow 2NO_{2( | vedclass

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(D) Let the initial pressure of $N_2O_5$ be $P_0$.The reaction is: $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$At $t=0$: $P_0$,$0$,$0$At

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$7/4$ times of initial pressure

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(D) Let the initial pressure of $N_2O_5$ be $P_0$.The reaction is: $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$At $t=0$: $P_0$,$0$,$0$At $t=t$: $P_0 - x$,$2x$,$\frac{1}{2}x$Total pressure $P_t = (P_0 - x) + 2x + \frac{1}{2}x = P_0 + \frac{3}{2}x$When $50\%$ of the reaction is completed,$x = \frac{P_0}{2}$.$P_t = P_0 + \frac{3}{2}(\frac{P_0}{2}) = P_0 + \frac{3}{4}P_0 = \frac{7}{4}P_0$.Thus,the final pressure is $7/4$ times the initial pressure.

For a reaction,$N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$ in a constant volume container,no products were present initially. The final pressure of the system when $50\%$ of the reaction is completed is:

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