A) 0.1
B) 0.4
C) 0.3
D) 0.2
Correct Answer: D
Solution :
\[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)\] Molar concentration of \[[{{N}_{2}}{{O}_{4}}]=\frac{9.2}{92}=0.1\,mol\text{/}L\] In equilibrium state when it 50% dissociates, \[[{{N}_{2}}{{O}_{4}}]=0.05\,M\] \[[N{{O}_{2}}]=0.1\,M\] \[{{K}_{c}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}\] \[=\frac{0.1\times 0.1}{0.05}=0.2\]You need to login to perform this action.
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