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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
  1. Consider the following endothermic reaction:
$\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
  1. Write expression for $K_p$ for the above reaction.
  2. How will the values of $K_p$ and composition of equilibrium mixture be affected by.
  1. increasing the pressure.
  2. increasing the temperature.
  3. using a catalyst?
  1. Calculate the pH of the resultant mixture of $10$ml of $0.1$M $H_2SO_4 + 10$ ml of $0.1$M KOH.
✓

Answer

  1. $\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
  1. $\text{K}_{\text{p}}=\frac{(\text{p}_{\text{CO}})(\text{p}_{\text{H}_2})^3}{(\text{p}_{\text{CH}_4})(\text{p}_{\text{H}_2\text{O}})}$
  2.  
  1. On increasing pressure, the reaction equilibria will shift in the backward direction.
  2. As the given reaction is endothermic, on increasing temperature the given equilibrium will shift in forward direction.
  3. There is no effect of catalyst in equilibrium composition, however the equilibrium will be attained faster.
  1. $10ml$ of $0.1M H_2SO_4$ is mixed with $10ml$ of $0.1$M KOH.
The reaction is
$2\text{KOH}+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{H}_2\text{O}$
$10$ ml of $0.1 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4=0.1 \times 10=1$ millimole
$10$ ml of $0.1 \mathrm{M} \mathrm{~KOH}=0.1 \times 10=1$ millimole
$1$ millimole of $KOH$ will react with $0.5$ millimole of $\mathrm{H}_2 \mathrm{SO}_4$
$\left[\because 0.5\right.$ millimole will produce 1 millimole of $\left.\mathrm{H}^{+}\right]$
$\therefore \mathrm{H}_2 \mathrm{SO}_4$ left $=1-0.5=0.5$ millimole
Volume of reaction mixtrue
$=10+10=20 \mathrm{ml}$
$\therefore$ Molarity of $x$ in the mixtrue.
$\therefore$ Molarity of × in the mixtrue.
$=\frac{0.5}{20}=2.5\times10^{-2}\text{M}$
$[\text{H}^+]=2\times2.5 \times10^{-2}$
$=5\times10^{-2}\text{M}$
$\text{pH}=-\log(5\times10^{-2})$
$=-\log5-\log10^{-2}$
$=-0.6990+2.0000=1.30$

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