Question

The rate constant for this first-order reaction is 0.053s ^ - 1 at 400 deg * c a products macmillan learning after how many seconds will 20.5 % of the reactant remain?

Answer 1

It will take approximately 54.45 seconds for 20.5% of the reactant to remain in a first-order reaction with a rate constant of 0.053 s^-1.

Step-by-step explanation:

To calculate how long it will take for 20.5% of a reactant to remain in a first-order reaction with a rate constant (k) of 0.053 s-1, we use the first-order integrated rate law, which is ln([A]/[A0]) = -kt, where [A] is the final concentration of the reactant, [A0] is the initial concentration, and t is the time.

Since we want 20.5% of the reactant to remain, [A]/[A0] = 0.205. To solve for t, rearrange the equation to t = -(1/k) ln([A]/[A0]), and substitute the given values:

t = -(1/0.053 s-1) ln(0.205)

t = -(-2.886)/(0.053 s-1)

t ≈ 54.45 s

So it will take approximately 54.45 seconds for 20.5% of the reactant to remain.