N_2O_5 decomposes to NO_2 and O_2 and follows | vedclass

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(B) The decomposition reaction is: $N_2O_5(g) 	o 2NO_2(g) + \frac{1}{2}O_2(g)$At $t = 0$: $P_{N_2O_5} = 50 \, mm \, Hg$,$P_{NO_2} = 0$,$P_{O_2} = 0$.

Vedclass · Accepted Answer

$106.25$

Vedclass · Answer

(B) The decomposition reaction is: $N_2O_5(g) 	o 2NO_2(g) + \frac{1}{2}O_2(g)$At $t = 0$: $P_{N_2O_5} = 50 \, mm \, Hg$,$P_{NO_2} = 0$,$P_{O_2} = 0$. Total pressure $P_0 = 50 \, mm \, Hg$.At $t = 50 \, min$: Let $p_1$ be the decrease in pressure of $N_2O_5$. The partial pressures are: $P_{N_2O_5} = 50 - p_1$,$P_{NO_2} = 2p_1$,$P_{O_2} = 0.5p_1$.Total pressure $P_t = (50 - p_1) + 2p_1 + 0.5p_1 = 50 + 1.5p_1 = 87.5 \, mm \, Hg$.$1.5p_1 = 37.5 \implies p_1 = 25 \, mm \, Hg$.Since $p_1 = 25$ is half of the initial pressure $50$,the half-life $t_{1/2} = 50 \, min$.At $t = 100 \, min$ $(2 	imes t_{1/2})$,the remaining pressure of $N_2O_5$ is $50 	imes (1/2)^2 = 12.5 \, mm \, Hg$.Thus,$50 - p_2 = 12.5 \implies p_2 = 37.5 \, mm \, Hg$.Total pressure at $100 \, min = 50 + 1.5p_2 = 50 + 1.5(37.5) = 50 + 56.25 = 106.25 \, mm \, Hg$.

$N_2O_5$ decomposes to $NO_2$ and $O_2$ and follows first order kinetics. After $50 \, min$,the pressure inside the vessel increases from $50 \, mm \, Hg$ to $87.5 \, mm \, Hg$. The pressure of the gaseous mixture after $100 \, min$ at constant temperature will be ........... $mm \, Hg$

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