A) 150 mm He
B) 116.05 mm Hg
C) 106.25 mm Hg
D) 125 mm Hg
Correct Answer: C
Solution :
First order reaction rate \[=K\left[ {{N}_{2}}{{O}_{5}} \right]\] \[2{{N}_{2}}{{O}_{5(g)}}\xrightarrow[{}]{{}}4N{{O}_{2(g)}}+{{O}_{2(g)}}\] t = 0 pressure 50 0 0 t = 30 t pressure 50- 2p 4p p Total pressure \[50-2p+4p+p=50+3p=87.5mmHg\] \[\therefore \] \[P=12.5mmHg\] \[\therefore \]\[{{P}_{0}}=50\And {{P}_{t}}=25\]for\[{{N}_{2}}{{O}_{5}}\]reactant \[\therefore \]\[K=\frac{2.303}{30\min }\times \log \left( \frac{50}{25} \right)=\frac{2.303}{60\min }\times \log \left( \frac{50}{x} \right)\]On solving \[x=12.5mmHg=50-2p\] \[\therefore \]\[P=18.75mmHg\] \[\therefore \]Total pressure\[=50+3p=106.25mmHg\]You need to login to perform this action.
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