The decomposition of NH₃ on a Pt surface is a zero-order reaction. If the value of the rate constant is 2 × 10⁻⁴ mole litre⁻¹ s⁻¹, what are the rates of appearance of N₂ and H₂ respectively?

Correct option is D. N₂=2 × 10⁻⁴ mol l⁻¹ s⁻¹, H₂= 6 × 10⁻⁴ mol⁻¹ s⁻¹

Solution:-
Rate of reaction for the decomposition of ammonia can be written as:
-d[NH₃]/dt = -(1/2)d[N₂]/dt = (1/3)d[H₂]/dt

Here, k = Rate constant
The reaction is of zero order.
Therefore,
Rate of appearance of N₂ = d[N₂]/dt = 2k = 2 × 2 × 10⁻⁴ mol lit⁻¹ s⁻¹ = 4 × 10⁻⁴ mol lit⁻¹ s⁻¹
Rate of appearance of H₂ = d[H₂]/dt = 3k = 3 × 2 × 10⁻⁴ = 6 × 10⁻⁴ mol lit⁻¹ s⁻¹

The provided solution in the input data is incorrect. The correct rates are derived above. There appears to be a mistake in the original solution's calculations and the given correct option. The correct option should reflect the calculated values. There was a 2 in the equation instead of a 1/2. Using the correct stoichiometry, the correct rates are 2k for N2 and 3k for H2. The correct answer should be D, but the values in option D are also incorrect. The correct answer should be N2=210^-4 mol L-1 s-1 and H2=610^-4 mol L-1 s-1