The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10

Question

The rate for the decomposition of NH3 on platinum surface is zero order. What are the rate of production of N2 and H2 if K=2.5×10-4mollitre-1s-1? Select the correct answer from above options

Answer 1

Correct Answer - d[N2] dt =1.25×10-4Ms-1 d[H2] dt =3.75×10-4Ms-1 2NH3→N2+3H2 Rate=k[NH3]0⇒k=2.5×10-4Ms-1 Rate -d[NH3] dt = 1 2 ( -d[NH3] dt ) = 1 2 ×2.5×10-4Ms-1 =1.25×10-4Ms-1 ∴ d[H2] dt = 3 2 ( -d[NH3] dt ) = 3 2 ×2.5×10-4Ms-1=3.75×10-4Ms-1