A) 28.34
B) 32.66
C) 34.78
D) 38.88
Correct Answer: B
Solution :
Given \[[{{H}_{2}}]=8.0\,mol/L\] \[[{{I}_{2}}]=3.0\,mol/L\] \[[HI]=28\,mol/L\] \[K=?\] \[{{H}_{2}}+{{I}_{2}}2HI\] \[\therefore \] \[K=\frac{[HI]}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{(28)}^{2}}}{(8)\times (3)}\] \[=\frac{28\times 28}{24}\] \[=32.66\]You need to login to perform this action.
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