A) \[3.465\times {{10}^{-2}}\,{{\min }^{-1}}\]
B) \[34.65\,{{\min }^{-1}}\]
C) \[3.465\,{{\min }^{-1}}\]
D) \[0.3465\,{{\min }^{-1}}\]
Correct Answer: A
Solution :
\[C{{H}_{3}}CHO(g)\to C{{H}_{4}}(g)+CO(g)\] \[\begin{matrix} \text{Initial} & 80\,mm\,Hg & 0 & {} \\ \text{Final} & 80-p & p & {} \\ \end{matrix}\begin{matrix} {} & 0 \\ {} & p \\ \end{matrix}\] After 20 min. \[80+p=120\] \[\Rightarrow \] \[p=40\,mm\,Hg\] i.e. 50% of reaction is complete. i.e., \[{{t}_{1/2}}\]= 20 min \[\therefore \] \[k=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{20}=3.465\times {{10}^{-2}}{{\min }^{-1}}\]
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