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(C) The reaction is $A(g) \longrightarrow B(g) + C(g)$.Let the initial pressure of $A$ be $P_i$.At $t = 0$: $P_A = P_i$,$P_B = 0$,$P_C = 0$. Total pre

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$80$

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(C) The reaction is $A(g) \longrightarrow B(g) + C(g)$.Let the initial pressure of $A$ be $P_i$.At $t = 0$: $P_A = P_i$,$P_B = 0$,$P_C = 0$. Total pressure $P_{total} = P_i = 180 \ mm \ of \ Hg$ (at completion).At $t = 20 \ min$: $P_A = P_i - x$,$P_B = x$,$P_C = x$.Total pressure $P_t = (P_i - x) + x + x = P_i + x = 100 \ mm \ of \ Hg$.Since $P_i = 180 \ mm \ of \ Hg$,we have $180 + x = 100$,which is not possible as $x$ must be positive. Re-evaluating the data: The total pressure at completion is $2P_i = 180 \ mm \ of \ Hg$,so $P_i = 90 \ mm \ of \ Hg$.At $t = 20 \ min$: $P_t = P_i + x = 100 \ mm \ of \ Hg$.$90 + x = 100 \implies x = 10 \ mm \ of \ Hg$.The partial pressure of $A$ at $20 \ min$ is $P_A = P_i - x = 90 - 10 = 80 \ mm \ of \ Hg$.

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