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In Contact process, SO3 is prepared by the oxidation of SO2 as per the following reaction: 2SO2 (g) + O2 (g) ⇌ 2SO3 (g); ∆H = -189.4a) What happens to the rate of forward reaction when (i) temperature is increased?(ii) pressure is decreased? (iii) a catalyst V2O5 is added? (b) Calculate the pH of 0.01 M H2SO4 solution. Also, calculate the hydroxyl ion concentration in the above solution. |
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Answer» (a) (i) When temperature is increased, the rate of forward reaction decreases since it is exothermic. (ii) When pressure is decreased the rate of forward reaction decreases since it is associated with decrease in number of moles. (iii) When a catalyst V2O5 is added the rate of both forward and backward reactions are increased by the same extent and equilibrium is reached earlier. (b) [H+] = 0.01M x 2 = 0.02M = 2 x 10-2M pH = -log[H+] = -log(2 x 10-2) = 1.6990 [OH-] = \(\frac{10^{-14}}{[H+]}\) = \(\frac{10^{-14}}{2\times10^{-2}}\) = 5 x 10-13 |
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