
The reaction $A(s) \to 2B(g) + C(g)$ is first order. The pressure after 20 min and after a very long time was 150mm Hg and 225nm Hg. The value of rate constant and pressure after 40 min are:
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
582.9k+ views
Hint: A first order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. The reaction rate or rate of reaction is the speed at which a chemical reaction takes place.
Formula used:
$k = \dfrac{{2.303}}{t}\log \dfrac{{{A_0}}}{{{A_t}}}$
Complete step by step answer:
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)
Hence option A is correct.
Note:
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 don’t hit each other as often and the rate of reaction decreases.
Formula used:
$k = \dfrac{{2.303}}{t}\log \dfrac{{{A_0}}}{{{A_t}}}$
Complete step by step answer:
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)
Hence option A is correct.
Note:
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 don’t hit each other as often and the rate of reaction decreases.
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