A) \[116.25\text{ mmHg}\]
B) \[106.25\text{ mmHg}\]
C) \[136.25\text{ mmHg}\]
D) \[175.0\text{ mmHg}\]
Correct Answer: B
Solution :
\[{{N}_{2}}{{O}_{5}}\to 2N{{O}_{2}}+\frac{1}{2}{{O}_{2}}\] \[R=k{{[{{N}_{2}}{{O}_{5}}]}^{1}}\] \[\begin{matrix} 1. & {{P}_{o}} & O & O & {} \\ 2. & {{P}_{o}}-P & 2p & \frac{1}{2}P & \to 50\min \\ 3. & {{P}_{o}}-{{p}^{1}} & 2{{p}^{1}} & \frac{1}{2}{{P}^{1}} & \to 100\min \\ \end{matrix}\] \[{{P}_{o}}=50\]\[\therefore {{P}_{50\min }}={{P}_{o}}+\frac{3}{2}P=87.5\] \[P=\frac{2}{3}(87.5-50)\] \[P=\frac{37.5}{3}\times 2=25\] First order \[E{{q}^{n}}:t=\frac{2.303}{k}\log \left[ \frac{({{N}_{2}}{{O}_{5}})}{{{({{N}_{2}}{{O}_{5}})}_{t}}} \right]\] At 50 min :\[t=\frac{2.303}{k}\log \] \[k=\frac{2.303}{50}\times 0.3010\] At 100min: \[100=\frac{2.303\times 50}{2.303\times 0.3010}\log \left[ \frac{{{({{N}_{2}}{{O}_{5}})}_{o}}}{{{({{N}_{2}}{{O}_{5}})}_{100}}} \right]\] \[2\times 0.3010={{\log }_{10}}\left[ \frac{50}{x} \right]\] \[4=\frac{50}{x}\] \[x=\frac{5025}{4}\] \[\therefore {{P}_{o}}-{{P}^{i}}=12.5\] \[{{P}^{1}}=50-12.5\] \[{{p}^{1}}=37.5\] \[\therefore \]Total Pressure\[={{P}^{o}}+\frac{3}{2}\times 37.5\] \[106.25mmHg\]
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