For the reaction A_{(g)} rightarrow B_{(g)} + | vedclass

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Question

(C) The reaction is $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$.At $t = 0$,pressure of $A = P_0 = 100 \ mm$,$P_B = 0$,$P_C = 0$. Total pressure $P_{total}

Vedclass · Accepted Answer

$\frac{2.303}{10} \log \frac{100}{80}$

Vedclass · Answer

(C) The reaction is $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$.At $t = 0$,pressure of $A = P_0 = 100 \ mm$,$P_B = 0$,$P_C = 0$. Total pressure $P_{total} = 100 \ mm$.At $t = 10 \ min$,let the pressure of $A$ reacted be $x$.$P_A = 100 - x$,$P_B = x$,$P_C = x$.Total pressure $P_t = (100 - x) + x + x = 100 + x = 120 \ mm$.So,$x = 20 \ mm$.The pressure of $A$ remaining at $t = 10 \ min$ is $P_A = 100 - 20 = 80 \ mm$.For a first-order reaction,$k = \frac{2.303}{t} \log \frac{P_0}{P_A}$.Substituting the values,$k = \frac{2.303}{10} \log \frac{100}{80} \ min^{-1}$.

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

For the reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$,the rate law is $R = k[A]$. At the start $(t = 0)$,the total pressure is $100 \ mm$ and after $t = 10 \ min$,the total pressure is $120 \ mm$. The rate constant $(min^{-1})$ is:

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