Question

The reaction: A -> B+C is known to be second order with respect to A and to have a rate constant of 0.225 M-1s-1 at 277 K. An experiment was run at this temperature where [A]o = 0.387 M. Calculate the concentration of B after 0.119 seconds has elapsed.

Answer 1

Given reaction,

A --------> B + C

Given that this reaction is second order w.r.t. A hence rate expression is given as,

Rate = k[A]².

Integrated rate law is given as,

[A] = [A]₀ / (1 + kt[A]₀) ........... (1)

Given data:

[A]₀ = 0.387 M, k = 0.225 M^-1.S^-1, t = 0.119 s.

On putting these values in equation (1)

[A] = (0.387) / (1 + 0.225 x 0.119 x 0.387)

[A] = (0.387) / (1.0103)

[A] = 0.383 M

[A] is the amount of reactant left after 0.119 s.

Amount of A reacted = [A]0 - [A] = 0.387 - 0.383 = 0.004 M

The given reaction is second order in A i.e. 2 moles of A react to form 1 mole of each of the product B andd C.

Hence after 0.119 s,

[B] = [A] / 2 = 0.004 / 2 = 0.002 M

Concentration of B formed after 0.119 s is 0.002 M.

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