A) 0.15 mol
B) 0.06 mol
C) 0.03 mol
D) 0.2 mol
Correct Answer: B
Solution :
Let\[x\]mol each of\[{{H}_{2}}\]and\[{{I}_{2}}\]react to give\[2x\]mole of\[HI\]at equilibrium. Then at equilibrium \[[{{H}_{2}}]=[{{I}_{2}}]=(0.3-x)/10\] \[\frac{{{[2n/10]}^{2}}}{\left[ \frac{0.3-x}{10} \right]\left[ \frac{0.3-x}{10} \right]}=\frac{4{{x}^{2}}}{{{(0.3-x)}^{2}}}\] Or \[8=\frac{2x}{0.3-x}=x=0.24\] Mole of\[{{I}_{2}}\]left unreacted\[=0.3-0.24=0.06\]You need to login to perform this action.
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