$\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
On increasing pressure, the reaction equilibria will shift in the backward direction.
As the given reaction is endothermic, on increasing temperature the given equilibrium will shift in forward direction.
There is no effect of catalyst in equilibrium composition, however the equilibrium will be attained faster.
The reaction is
$2\text{KOH}+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{H}_2\text{O}$
10ml of 0.1M H2SO4 = 0.1 × 10 = 1 millimole
10ml of 0.1M KOH = 0.1 × 10 = 1 millimole
1 millimole of KOH will react with 0.5 millimole of H2SO4
$[\because$ 0.5 millimole will produce 1 millimole of H+]
$\therefore$ H2SO4 left = 1 - 0.5 = 0.5 millimole
Volume of reaction mixtrue
= 10 + 10 = 20ml
$\therefore$ 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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