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Question

Answers

A.$0.05\ln 1.5{\min ^{ - 1}},200mm$

B.$0.05\ln 2{\min ^{ - 1}},300mm$

C.$0.05\ln 3{\min ^{ - 1}},300mm$

D.$0.05\ln 3{\min ^{ - 1}},200mm$

Answer
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$k = \dfrac{{2.303}}{t}\log \dfrac{{{A_0}}}{{{A_t}}}$

Now according to the question,

After 20 min,

$k = \dfrac{{2.303}}{{20}}\log \dfrac{{225}}{{75}}$

Now,

$k = 0.05$

The pressure after 40 min will be,

Using the formula

$0.05 = \dfrac{{2.303}}{{40}}(\log 225 - \log (225 - x))$

$0.8685 - \log 225 = - \log y$

\[0.8684 - 2.3522 = - \log y\]

$y = anti\log (1.4838)$

${[A]_t} = 200mm$ (Taken antilog)

When the pressure increases the molecules have less space in which they can move. Hence greater density of molecules increases the number of collisions. When the pressure decreases molecules donâ€™t hit each other as often and the rate of reaction decreases.