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Write a relation between `Delta G and Q` and define the meaning of each term and answer the following: (a) Why a reaction proceeds forward when `Q lt K` and no net reaction occurs when Q = K. (b) Explain the effect of increase in pressure in terms of reaction Q, for the reaction : `CO(g) + 3 H_(2) (g) hArr CH_(4)(g) + H_(2) O (g)`. |
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Answer» `Delta G = DeltaG^(@)` + RT ln Q where `Delta G =` Change in free energy as the reaction proceeds `Delta G^(@) = ` Standard free energy change Q = Reaction quotient R = Gas constant and T= temperature in K (a) As `Delta G^(@) = - ` RT ln K `:. Delta G = - RT` ln K + RT ln Q = RT ln `(Q)/(K)` If `Q lt K, Delta G` will be -ve. Reaction will proceed in the forward direction. If `Q=K, DeltaG=0`. Reaction is in equilibrium and there is no net reaction . (b) `K_(c)=([CH_(4)][H_(2)O])/([CO][H_(2)]^(3))`. If pressure is increased, volume decreases. Suppose presure is doubled, volume will be halved. Hence, molar concentrations will be doubled. Now `Q_(c) = ({2[CH_(4)]}{2[H_(2)O]})/({2[CO]}{2[H_(2)]}^(3))=(1[CH_(4)][H_(2)O])/(4[CO][H_(2)]^(3))=(1)/(4) K_(c)` This, `Q_(c)` is less than `K_(c)`. Hence, to re-establish equilibrium , `Q_(c) ` will tend to increase , i.e., equilibrium will shift in the forward direction. |
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