2 NO(g) + Br2(g) → 2 NOBr(g) NO(g) reacts with Br2(g), as represented by the equation above. An expe — Kinetics Chemistry Question
Problem Context
2 NO(g) + Br2(g) → 2 NOBr(g)
NO(g) reacts with Br2(g), as represented by the equation above. An experiment was performed to study the rate of the reaction at 546 K. Data from three trials:
Trial 1: [NO]=0.10 M, [Br2]=0.20 M, initial rate of consumption of Br2 = 12.0 M/s
Trial 2: [NO]=0.40 M, [Br2]=0.20 M, rate = 192.0 M/s
Trial 3: [NO]=0.10 M, [Br2]=0.60 M, rate = 36.0 M/s
Model Answer
1st order; tripling [Br2] triples the rate of reaction. One point is earned for correct order with justification. || 2nd order. Quadrupling [NO] increases the rate by 16x. One point is earned for correct order with justification.
Model Answer
Rate = k[NO]²[Br2]. One point is earned for the correct equation.
Model Answer
12.0 M/s = k(0.10)²(0.20); k = 6.0×10³ M⁻²s⁻¹. One point is earned for setup. One point is earned for correct numerical result with correct units.
Model Answer
[Br2]reacting = 0.20 M - 0.16 M = 0.040 M. [NO] = 0.40 M - 2(0.04 M)(1 mol NO/1 mol Br2) = 0.32 M. One point is earned for correct numerical result. || r = (6.0×10³ M⁻²s⁻¹)(0.32 M)²(0.16 M) = 98 M/s. One point is earned for correct numerical result.
Model Answer
No. The rate law for the proposed mechanism would be 1st order in NO, but the experimental rate law is 2nd order in NO. One point is earned for correct response with explanation.