Azo isopropane decomposes according to the eq | vedclass

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(A) For the reaction: $((CH_3)_2CHN)_2N_{2(g)} \rightarrow N_{2(g)} + C_6H_{14(g)}$Let the initial pressure of the reactant be $P_o$ at $t = 0$.At tim

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

Vedclass · Answer

(A) For the reaction: $((CH_3)_2CHN)_2N_{2(g)} \rightarrow N_{2(g)} + C_6H_{14(g)}$Let the initial pressure of the reactant be $P_o$ at $t = 0$.At time $t$,let the pressure of the reactant decomposed be $x$.Initial: $P_o, 0, 0$At time $t$: $(P_o - x), x, x$The total pressure $P_t = (P_o - x) + x + x = P_o + x$.Therefore,$x = P_t - P_o$.The pressure of the reactant remaining at time $t$ is $P_{reactant} = P_o - x = P_o - (P_t - P_o) = 2P_o - P_t$.For a first-order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log \frac{P_{initial}}{P_{remaining}}$$K = \frac{2.303}{t} \log \frac{P_o}{2P_o - P_t}$

Time	$t$	$\infty$
$P_{\text{system}}$	$P_t$	$P_{\infty}$

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