Decomposition of A is a first order reaction | vedclass

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(B) The reaction is $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$.Initially $(t=0)$: $P_A = 1 \ bar$, $P_B = 0$, $P_C = 0$. Total pressure $P_0 = 1 \ bar$.A

Vedclass · Accepted Answer

$6.9 	imes 10^{-3}$

Vedclass · Answer

(B) The reaction is $A_{(g)} ightarrow B_{(g)} + C_{(g)}$.Initially $(t=0)$: $P_A = 1 \ bar$, $P_B = 0$, $P_C = 0$. Total pressure $P_0 = 1 \ bar$.At time $t = 100 \ min$: $P_A = 1 - x$, $P_B = x$, $P_C = x$.Total pressure $P_t = (1 - x) + x + x = 1 + x$.Given $P_t = 1.5 \ bar$, so $1 + x = 1.5 \implies x = 0.5 \ bar$.Partial pressure of $A$ at $t = 100 \ min$ is $P_A = 1 - 0.5 = 0.5 \ bar$.For a first-order reaction, $k = \frac{2.303}{t} \log \frac{P_0}{P_A}$.$k = \frac{2.303}{100} \log \frac{1}{0.5} = \frac{2.303}{100} \log 2$.Using $\log 2 = 0.3$, $k = \frac{2.303 	imes 0.3}{100} = \frac{0.6909}{100} \approx 6.9 	imes 10^{-3} \ min^{-1}$.

Time $(t)$ $\text{min}$	$0$	$30$	$60$	$90$
Conc. of ester $(C)$ $\text{mol L}^{-1}$	$0.850$	$0.800$	$0.754$	$0.710$

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