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The rate constant for the reaction is 0.230 M − 1 ⋅ s − 1 0.230 M−1⋅s−1 at 200 ∘ C. 200 ∘C. A ⟶ products A⟶products If the initial concentration of A A is 0.00290 M, 0.00290 M, what will be the concentration after 155 s?

Respuesta :

Answer:

The concentration after 155 s = 0.00263 M

Explanation:

From the units of the reactions rate constant, K = 0.230 M⁻¹s⁻¹, it is evident that the reaction is second order with respect to the reactant A.

Let C = concentration of A At any time

And C₀ = initial concentration of A = 0.00290 M

dC/dt = - KC²

dC/C² = - kdt

C⁻² dC = - k dt

∫ C⁻² dC = -k ∫ dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

- (C⁻¹ - C₀⁻¹) = - kt

(C⁻¹ - C₀⁻¹) = kt

(1/C) - (1/C₀) = kt

t = 155 s, C₀ = 0.00290 M, k = 0.230 M⁻¹s⁻¹

(1/C) - (1/0.0029) = 0.23 × 155

(1/C) = 35.65 + 344.83 = 380.478

C = 1/380.478 = 0.00263 M

Histrionicus

The concentration after 155 s in the given case is - 0.00263 M

The units for the rate constant are given as M⁻¹⋅s⁻,¹ which corresponds to the units for the rate constant of a second order reaction.

Let A = concentration of A at any time

And A₀ = initial concentration of A = 0.00290 M

  • This is 2nd order and for a second order reaction, the integrated rate law is: 1/[A] = kt + 1/[A]o

So, for this reaction,

1/A = 1/A₀ + kt

substitute all values

1/A = 1/0.0029 + 0.23 * 155

(1/A) - (1/0.0029) = 0.23 × 155

(1/A) = 35.65 + 344.83 = 380.478

then, A = 1/380.478 = 0.00263 M

A =0.00263 M

Thus, the correct answer would be - 0.00263 M.

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