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Decomposition of NH₃ on gold surface follows zero order kinetics. If rate constant K is $5 \times 10^{- 4} M - s^{- 1}, rate of formation of N_{2} will be$

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10^{- 3} M - s^{- 1}

2.5 \times 10^{- 4} M - s^{- 1}

5 \times 10^{- 4} M - s^{- 1}

Zero

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detailed solution

Correct option is C

N₂+3H₂→2NH₃

K =5 $\times$ 10^-4M s−1

Rate = $- \frac{1 d [{NH}_{3}] }{2 dt}$ = $\frac{d [N_{2}] }{dt}$ = $\frac{1 d [{NH}_{3}] }{3 dt}$

Rate of formation of N₂ = $\frac{d [N_{2}] }{dt}$ = K =5 $\times$ 10^-4M s⁻¹

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Decomposition of NH3 on gold surface follows zero order kinetics. If rate constant K is 5×10−4M−s−1, rate of formation of N2 will be

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Decomposition of NH₃ on gold surface follows zero order kinetics. If rate constant K is $5 \times 10^{- 4} M - s^{- 1}, rate of formation of N_{2} will be$