A) \[1.15atm\]
B) \[1.32atm\]
C) \[1.22atm\]
D) \[1.12atm\]
Correct Answer: D
Solution :
: \[{{t}_{1/2}}=69.3s,\,\,t=230s,\,\,a=0.4atm\] For first order reaction, \[k=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{69.3}=0.01{{s}^{-1}}\] \[k=\frac{2.303}{t}\log \frac{a}{a-x}=\frac{2.303}{230}\log \frac{0.4}{0.4-x}\] \[0.01=\frac{2.303}{230}\log \frac{0.4}{0.4-x}=0.01\log \frac{0.4}{0.4-x}\] \[\frac{0.01}{0.01}=\log \frac{0.4}{0.4-x}\] \[\Rightarrow \] \[10=\frac{0.4}{0.4-x}\] \[x=0.36\] For the given reaction, \[\begin{matrix} {} & {{A}_{(g)}}\xrightarrow{\Delta } & {{P}_{(g)}} & {{Q}_{(g)}} & {{R}_{(g)}} \\ Initial & 0.4 & 0 & 0 & 0 \\ Final & 0.4-0.36 & 0.36 & 0.36 & 0.36 \\ \end{matrix}\] Total pressure \[=(0.4-0.36)+(3\times 0.36)=1.12\text{ }atm\]You need to login to perform this action.
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