| S.no. | Time | Total pressure in Pascal |
| 1. | At the end of 10 min | 300 |
| 2. | After completion | 200 |
A) 0.0693
B) 69.3
C) 6.93
D) \[6.93\times {{10}^{-4}}\]
Correct Answer: A
Solution :
\[\underset{2-2x}{\mathop{2A\,(g)}}\,\xrightarrow{{}}\,\underset{x}{\mathop{B\,(g)}}\,+C\,(s)\] At the end of reaction, only 1 mole of gas is present whose pressure is 200 Pascal. \[\therefore \] At the beginning of the reaction 2 moles of gas should have a pressure of 400 Pascal. After time 10 min No. of moles present, \[2-2x+\text{ }x=2-x\] The pressure of 2 moles = 400 \[\therefore \] \[400-x=300\] \[\therefore \] \[x=100\] \[\therefore \] Pressure due to \[2-2x\] moles of A \[=400-200=200\] \[\therefore \] \[k=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)=\frac{2.303}{10}\log \left( \frac{400}{200} \right)\] \[=\frac{2.303}{10}\log \,\,2=\frac{0.693}{10}\] \[=0.0693\,{{\min }^{-1}}\]
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