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InterviewSolution
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In a reaction between `H_(2) and I_(2)` at a certain temperature, the amounts of `H_(2), I_(2) and HI` at equilibrium constant f werefound to be 0.45 mole, 0.39 mole and 3.0 moles respectively. Calculate the equilibrium constant for the reaction at the given temperature. |
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Answer» Thr reaction between `H_(2) and I_(2)` may be represented as : `H_(2) + I_(2) hArr 2 HI` Amounts of `H_(2), I_(2) and HI` at equilibrium are given to be `H_(2) = 0.45 " mole " , I_(2) = 0.39 " mole" and HI= 3.0 " mole"` Suppose the volume of the vessel ( i.e., reaction mixture) = V litres. Then the molar concentrations at equilibrium will be `[H_(2)] = (0.45)/V molL^(-1), [I_(2)]= (0.39)/V mol L^(-1) and [ HI] = (3.0)/V mol L^(-1)` Applying the law of chemical equilibrium to the above reaction, we get `K_(c) = [HI]^(2) /([H_(2)][I_(2)])=(3.0//V)^(2)/(0.45 xx 0.39) = 51.28` |
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