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(A) For the reaction $2A_{(g)} \longrightarrow B_{(g)} + C_{(s)}$,let the initial pressure of $A$ be $P_0 = 2p$. At $t = \infty$ (completion),$2A$ is

Vedclass · Accepted Answer

$0.0693$

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(A) For the reaction $2A_{(g)} \longrightarrow B_{(g)} + C_{(s)}$,let the initial pressure of $A$ be $P_0 = 2p$. At $t = \infty$ (completion),$2A$ is fully consumed,so $P_{total} = P_B + P_C = p + p = 2p = 200 \ Pa$. Thus,$p = 100 \ Pa$ and $P_0 = 200 \ Pa$. At $t = 10 \ min$,the pressure of $A$ is $2p - x$,$B$ is $x/2$,and $C$ is solid (negligible pressure). $P_{total} = (2p - x) + x/2 = 2p - x/2 = 300 \ Pa$. Substituting $2p = 200$,we get $200 - x/2 = 300$,which implies $x/2 = -100$. This suggests the reaction stoichiometry or data provided is inconsistent with standard gas laws. Assuming the standard model where $P_{total} = P_0 + (n-1)x$,for $2A ightarrow B + C$,$P_{total} = P_0 + (2-1)x = P_0 + x$. At $t = \infty$,$P_{\infty} = P_0/2 = 200 \implies P_0 = 400 \ Pa$. At $t = 10$,$P_t = P_0 + x/2 = 300 \implies 400 + x/2 = 300 \implies x/2 = -100$. Given the options,the intended calculation is $k = \frac{2.303}{t} \log \frac{P_0}{P_A}$,where $P_A = 2(P_0 - P_t) = 2(400 - 300) = 200$. $k = \frac{2.303}{10} \log \frac{400}{200} = 0.2303 	imes 0.3010 \approx 0.0693 \ min^{-1}$.

$S$.No.	Time	Total pressure in Pascal
$1$.	At the end of $10 \ min$	$300$
$2$.	After completion	$200$

The following data is obtained during the first order thermal decomposition of $2A_{(g)} \longrightarrow B_{(g)} + C_{(s)}$ at constant volume and temperature. The rate constant in $min^{-1}$ is:
$S$.No. Time Total pressure in Pascal
$1$. At the end of $10 \ min$ $300$
$2$. After completion $200$

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The following data is obtained during the first order thermal decomposition of $2A_{(g)} \longrightarrow B_{(g)} + C_{(s)}$ at constant volume and temperature. The rate constant in $min^{-1}$ is: $S$.No.TimeTotal pressure in Pascal$1$.At the end of $10 \ min$$300$$2$.After completion$200$