A) \[5.2 \times 1{{0}^{-}}^{4}se{{c}^{-}}^{1}; \,0.05 atm\]
B) \[6.2 \times 1{{0}^{-}}^{3}se{{c}^{-}}^{1}; \,0.033 atm\]
C) \[5.8 \times 1{{0}^{-}}^{3}se{{c}^{-}}^{1}; \,0.44 atm\]
D) \[4.6 \times 1{{0}^{-}}^{3}se{{c}^{-}}^{1}; \,0.005 atm\]
Correct Answer: B
Solution :
The equation for first-order reaction is \[6.20\times {{10}^{-3}}=\frac{2.303}{t}log\,\left( \frac{a}{a-x} \right)\] (i) Here \[\operatorname{a}= 0.062;\,\,a - x =0.044\] \[k=\frac{2.303}{55}\log \,\frac{0.062}{0.044}=6.2\times {{10}^{-3}}\,{{\sec }^{-1}}\] (ii) Here we are to find the value of \[\operatorname{a} - x\] \[6.20\times {{10}^{-3}}=\frac{2.303}{100}\log \frac{0.062}{a-x}\] \[\log \frac{0.062}{a-x}=2.69\times {{10}^{-1}}\] \[\frac{0.062}{a-x}=1.86\] \[a-x=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)=0.033\,\,atm\]
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