The reaction described by the equation O 3 (g) + NO (g) ⟶ O 2 (g) + NO 2 (g) has, at 310 K, the rate law rate of reaction = k [ O 3 ] [ NO ] k = 3.0 * 10 6 M - 1 ⋅ s - 1 Given that [ O 3 ] = 7.0 * 10 - 4 M and [ NO ] = 5.0 * 10 - 5 M at t = 0, calculate the rate of the reaction at t = 0.

Question

Derek Brennan

Chemistry

4 December, 04:41

The reaction described by the equation O 3 (g) + NO (g) ⟶ O 2 (g) + NO 2 (g) has, at 310 K, the rate law rate of reaction = k [ O 3 ] [ NO ] k = 3.0 * 10 6 M - 1 ⋅ s - 1 Given that [ O 3 ] = 7.0 * 10 - 4 M and [ NO ] = 5.0 * 10 - 5 M at t = 0, calculate the rate of the reaction at t = 0.

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Ada Burton · Answer 1

Equation of the reaction:

O3 (g) + NO (g) ⟶ O2 (g) + NO2 (g)

From the Rate Law:

R = k[O3][NO]

The order of NO is 1 and that of O3 is also 1, therefore order of reaction = 1 + 1

= 2

Given:

Concentration of O3, [O3] = 7.0*10^-4M

Concentration of NO, [NO] = 5.0*10^-5M

Rate constant, k = 3.0 * 10^6M^-1s^-1

R = 3.0*10^6 * 7.0*10^-4 * 5.0*10^-5

= 0.105 M/s