Azo-isopropane decomposes according to the fo | vedclass

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(A) For the reaction: $((CH_3)_2CHN = NCH(CH_3)_2)_{(g)} \rightarrow (N_2)_{(g)} + (C_6H_{14})_{(g)}$At $t=0$: Initial pressure = $P_o$,$N_2$ = $0$,$C

Vedclass · Accepted Answer

$K = \frac{2.303}{t} \log \frac{P_o}{2P_o - P_t}$

Vedclass · Answer

(A) For the reaction: $((CH_3)_2CHN = NCH(CH_3)_2)_{(g)} \rightarrow (N_2)_{(g)} + (C_6H_{14})_{(g)}$At $t=0$: Initial pressure = $P_o$,$N_2$ = $0$,$C_6H_{14}$ = $0$At time $t$: Pressure = $(P_o - x)$,$N_2$ = $x$,$C_6H_{14}$ = $x$The total pressure $P_t$ at time $t$ is given by:$P_t = (P_o - x) + x + x = P_o + x$Therefore,$x = P_t - P_o$The partial pressure of the reactant at time $t$ is $(P_o - x) = P_o - (P_t - P_o) = 2P_o - P_t$For a first-order reaction,the rate constant $K$ is:$K = \frac{2.303}{t} \log \frac{P_o}{P_o - x}$Substituting $(P_o - x) = 2P_o - P_t$:$K = \frac{2.303}{t} \log \frac{P_o}{2P_o - P_t}$

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