Chemical Kinetics and Nuclear ChemistryHard
Question
The rate constant for the reaction, 2N2O5 → 4NO2 + O2 is 3.0 × 10-4s-1. If start made with 1.0 mol L-1 of N2O5, calculate the rate of formation of NO2 at the moment of the reaction ehen concentration of O2 is 0.1 mol L-1.
Options
A.2.7 × 10-4 molL-1s-1
B.2.4 × 10-4 molL-1s-1
C.4.8 × 10-4 molL-1s-1
D.9.6 × 10-4 molL-1s-1
Solution
Mol L-1 of N2O2 reacted = 2 × 0.1 = 0.2
[N2O5] left - 1.0 - 0.2 = 0.8 mol-1
Rate of reaction = k × [N2O5]
= 3.0 × 10-4 × 0.8
= 2.4 × 10-4 mol L-1s-1
Rate of formation of NO2
= 4 × 2.4 × 10-4 = 9.6 × 10-4 molL-1s-1
[N2O5] left - 1.0 - 0.2 = 0.8 mol-1
Rate of reaction = k × [N2O5]
= 3.0 × 10-4 × 0.8
= 2.4 × 10-4 mol L-1s-1
Rate of formation of NO2
= 4 × 2.4 × 10-4 = 9.6 × 10-4 molL-1s-1
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