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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
  1. For the reaction:
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}),$
The value of $K_p$ is $3.6 \times 10^{-2}$ at $500K$.
Colculate the value of $K_c$ for the reaction at the same temperature R = 0.083L bar $K^{-1}mol^{-1}$​​​​​​​.
  1. What is the effect of increasing pressure in the reactions? Give reason.
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2\text{(g)}$
$\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO(g)}$
✓

Answer

  1. The reaction is
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$
Given: $\text{K}_\text{p}=3.6\times10^{-2}$ at 500K
The reation between $K_p$ and $K_c$ is
$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$
For the above reaction
$\Delta \text{n}=2-4=-2$
$\text{K}_\text{c}=\frac{\text{K}_\text{p}}{(\text{RT})^{\Delta\text{n}}}$ (R = 0.083bar L $K^{-1}mol^{-1}$​​​​​​​)
$=\frac{3.6\times10^{-2}}{(0.083\times500)^{-2}}$
$=3.6\times10^{-2}\times(0.083\times500)^2$
$=62$​​​​​​​
  1.  
  2. The equilibrium will shift in backward reaction because number of moles of products are more than reactants $\Delta \text{n}>0.$
  3. No effect because number of moles of reactants and products are equal, i.e., $\Delta \text{n}=0.$

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