A) 1.2 atm
B) 2.4 atm
C) 2.0atm
D) 1.0 atm
Correct Answer: B
Solution :
\[{{N}_{2}}{{O}_{4}}(g)\] | \[\rightleftharpoons \] | \[2N{{O}_{2}}(g)\] | |
At start | 100/92mol=1.09mol | 0 | |
At equilibrium | 80/92mol=0.87mol | 20/46mol=0.43mol |
According to ideal gas equation, at two conditions |
At 300K; \[{{P}_{1}}V={{n}_{1}}R{{T}_{1}}\] |
\[1\times V=1.09\times R\times 300..(i)\] |
At 600K; \[{{P}_{2}}V={{n}_{2}}R{{T}_{2}}\] |
\[{{P}_{2}}\times V=\left( 0.87+0.43 \right)\times R\times 600..(ii)\] |
Divide (ii) by (i) |
\[\frac{{{P}_{2}}}{1}=\frac{1.30\times 600}{1.09\times 300};\] \[{{P}_{2}}\frac{1.30\times 2}{1.09}=2.39\operatorname{atm}.\simeq 2.4\operatorname{atm}.\] |
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