A) 0.33 atm
B) 1.33 atm
C) 2.33 atm
D) 3.33 atm
Correct Answer: B
Solution :
\[{{N}_{2}}{{O}_{4}}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,2N{{O}_{2}}(g)\] At. equil. 1 mol 2 mol Total moles = 1 + 2 = 3 \[\therefore \] \[{{n}_{{{N}_{2}}{{O}_{4}}}}=\frac{1}{3}\] and \[{{n}_{N{{O}_{2}}}}=\frac{2}{3}\] \[\because \] \[{{p}_{{{N}_{2}}{{O}_{4}}}}={{n}_{{{N}_{2}}{{O}_{4}}}}\times p\] \[\therefore \] \[{{p}_{{{N}_{2}}{{O}_{4}}}}=\frac{1}{3}\times 1=\frac{1}{3}\] and \[{{p}_{N{{O}_{2}}}}=\frac{2}{3}\times 1=\frac{2}{3}\] \[{{K}_{p}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{{{p}_{{{N}_{2}}{{O}_{4}}}}}\] \[=\frac{{{\left( \frac{2}{3} \right)}^{2}}}{\left( \frac{1}{3} \right)}\] \[=\frac{4}{3}=1.33\,atm\]
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