4.12 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1?
4.12 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1?
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1 Answer
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4.12 2NH3 (g)? N2 (g) + 3H2 (g)
Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 * 10-4mol L-1sec-1.
According to rate law,
-d [NH3] / 2dt = d [N2] / dt
2.5 * 10-4mol L-1sec-1 = d [N2] / dt
i.e. the rate of production of N2 is 2.5 * 10-4mol L-1 sec-1.
According to rate law,
-d [NH3] / 2dt = d [H2] / 3dt
d [H2] / dt = -3 X d [NH3] / 2dt
i.e. rate of formation of H2 is 3 times rate of reaction = 3 * 2.5 * 10-4mol L-1sec-1
= 7.5 * 10-4mol L-1sec-1
Rate of formation of N2 and H2 is 2.5 * 10-4 mol L-1sec-1 and 7.5 * 10-4 mol L-1sec-1 respectively
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