Difficult
The reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?
- ARate $= k[Z]$
- BRate $= k[X][Y]^2[Z]$
- CRate $= k[N]$
- DRate $= k[X][Y][Z]$
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Similar Questions
For the reaction $NH_4^+ + NO_2^- \to N_2 + 2H_2O$,the experimental data is given below. Determine the rate law for the reaction.
$1.$ $[NH_4^+] = 0.24 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 7.2 \times 10^{-6} \, M/s$ $2.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 3.6 \times 10^{-6} \, M/s$ $3.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.15 \, M, \text{Rate} = 5.4 \times 10^{-6} \, M/s$
| $1.$ $[NH_4^+] = 0.24 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 7.2 \times 10^{-6} \, M/s$ |
| $2.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 3.6 \times 10^{-6} \, M/s$ |
| $3.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.15 \, M, \text{Rate} = 5.4 \times 10^{-6} \, M/s$ |
Medium
View SolutionSelect the incorrect option :
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MediumIIT 2013
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