A) 0.1
B) 0.2
C) 0.4
D) 2
Correct Answer: B
Solution :
[b] 0.2 Initial \[[{{N}_{2}}{{O}_{4}}]=\frac{9.2}{92}\,mol{{L}^{-1}}\] \[=0.1\text{ }mol{{L}^{-1}}\] At equilibrium (after 50% dissociation) \[[{{N}_{2}}{{O}_{4}}]=0.05M,[N{{O}_{2}}]=0.1M\] \[K=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}\] \[K=\frac{{{[0.1]}^{2}}}{0.05}=0.2\]You need to login to perform this action.
You will be redirected in
3 sec