The reaction 2N_2O_5 ,(g) to 4NO_2 ,(g) + O_2 | vedclass

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(A) For the reaction $2N_2O_5 \,(g) 	o 4NO_2 \,(g) + O_2 \,(g)$,let the initial pressure of $N_2O_5$ be $P_0 = 50 \, mmHg$.At $t = 30 \, min$,let the

Vedclass · Accepted Answer

$106.25$

Vedclass · Answer

(A) For the reaction $2N_2O_5 \,(g) 	o 4NO_2 \,(g) + O_2 \,(g)$,let the initial pressure of $N_2O_5$ be $P_0 = 50 \, mmHg$.At $t = 30 \, min$,let the pressure of $N_2O_5$ decrease by $2p$. The total pressure is $P_t = (50 - 2p) + 4p + p = 50 + 3p = 87.5 \, mmHg$.Thus,$3p = 37.5 \, mmHg$,so $p = 12.5 \, mmHg$.The pressure of $N_2O_5$ remaining at $t = 30 \, min$ is $50 - 2(12.5) = 25 \, mmHg$.Since the pressure of $N_2O_5$ halves in $30 \, min$,the half-life $t_{1/2} = 30 \, min$.At $t = 60 \, min$ (which is $2 	imes t_{1/2}$),the pressure of $N_2O_5$ remaining is $50 / 4 = 12.5 \, mmHg$.Let $50 - 2p' = 12.5$,so $2p' = 37.5$,which means $p' = 18.75 \, mmHg$.The total pressure at $t = 60 \, min$ is $50 + 3p' = 50 + 3(18.75) = 50 + 56.25 = 106.25 \, mmHg$.

$S.NO$	$Time/s$	Total pressure $(atm)$
$1.$	$0$	$0.6$
$2.$	$100$	$X$

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