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The following reaction has attained equilibrium `CO(g) + 2 H_(2) (g) hArr CH_(3)OH (g) , Delta H^(@) = - 92*0 " KJ mol"^(-1)` What will happen if (i) volume of the reaction vessel is suddenly reduced to half ? (ii) the partial pressure of hydrogen is suddenly doubled ? (iii) an inert gas is added to the system ? |
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Answer» `K_(c) = ([ CH_(3)OH])/([CO][H_(2)]^(2)),K_(p)=(p_(CH_(3)OH))/(p_(CO)xx p_(H_(2))^(2))` (i) When volume of the vessel is reduced to Half, the concentration of each reactant or product becomes double. Thus, `Q_(c) = (2[CH_(3)OH])/(2[CO]xx{2[H_(2)]}^(2))=1/4 K_(c)` As ` Q_(c) lt K_(c)`, equilibrium will shift in the forward direction, producing more of `CH_(3)OH " to make " Q_(c) = K_(c)`. (ii) ` Q_(p) = (p_(CH_(3)OH))/p_(CO) xx (2 p_(H_(2)))^(2) = 1/4 K_(p).` Again , `Q_(p) lt K_(p).` equilibrium will shift in the forward direction to make `Q_(p) = K_(p)`. (iii) As volume remains constant, molar concentrations will not change. Hence there is no effect on the state of equilibrium. |
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