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 \[[{{N}_{2}}{{O}_{4}}]=\frac{9.2}{92}=0.1\,\,mol/L\] 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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