AIIMS 2017
- Exam
- AIIMS
- Year
- 2017
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
AIIMS 2017 · Chemistry · Chemical Equilibrium
For the reaction, H_2+ I _2 leftharpoons 2 HI, K=47.6 . If the...
For the reaction, \mathrm{I}. If the initial number of moles of each reactant and product is 1 mole then at equilibrium
Options
- A\left[\mathrm{I}_2\right]=\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]>[\mathrm{HI}]
- B\left[\mathrm{I}_2\right]=\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right] < [\mathrm{HI}]
- C\left[\mathrm{I}_2\right] < \left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]=[\mathrm{HI}]
- D\left[\mathrm{I}_2\right]>\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]=[\mathrm{HI}]
Correct answer
B. \left[\mathrm{I}_2\right]=\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right] < [\mathrm{HI}]
Step-by-step solution
For the given reaction, K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
As 1 mole of reacts with 1 mole of , even at equilibrium, \left[\mathrm{H}_2\right]=\left[\mathrm{I}_2\right]
Hence, K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{I}_2\right]^2} \quad or \quad \sqrt{K}=\frac{[\mathrm{HI}]}{\left[\mathrm{I}_2\right]}=\sqrt{47.6} i.e., [\mathrm{HI}]>\left[\mathrm{I}_2\right]