Answer:
(a) \[{{K}_{p}}=\frac{[{{p}_{CO}}]{{[{{p}_{{{H}_{2}}}}]}^{3}}}{[{{p}_{C{{H}_{4}}}}][{{p}_{{{H}_{2}}O}}]}\]
(b) (i) \[\Delta
{{n}_{g}}=4-2=2\]
\[\Delta
{{n}_{g}}>0\]
The reaction will
shift in backward direction (lower volume direction) when pressure is
increased. However, \[{{K}_{p}}\]is unaffected by \[{{K}_{p}}\]pressure change.
(ii) The
reaction is endothermic, hence, the equilibrium will shift in forward direction
by rise in temperature. The equilibrium constant \[{{K}_{p}}\] will also
increase with rise in temperature.
\[{{\log
}_{10}}\left( \frac{{{K}_{2}}}{{{K}_{1}}} \right)=\frac{\Delta H}{2.303}\left[
\frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
(iii) Composition of equilibrium mixture and the value of \[{{K}_{p}}\],
both are unaffected by the catalyst.
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