For the first-order reaction below,the concentration of product B after 24.2 seconds is 0.322 M.
If k = 8.75 × 10−² s−¹,what was the initial concentration of A? A → 2B
Rate = k[A]

Question

For the first-order reaction below,the concentration of product B after 24.2 seconds is 0.322 M. If k = 8.75 × 10−² s−¹,what was the initial concentration of A? A → 2B Rate = k[A]

Quizplus · Accepted Answer

We can use the integrated rate law for a first-order reaction:

ln([A]0/[A]) = kt

We can rearrange this equation to solve for [A]0:

[A]0 = [A] / e^(kt)

We are given that after 24.2 seconds, [B] = 0.322 M. We can use the stoichiometry of the reaction to relate this to [A]:

2 moles of B are produced for every 1 mole of A that reacts.

Thus, [B] = 2[A]0 - 2[A]

Substituting the given values, we get:

0.322 M = 2[A]0 - 2[A]

We need to know the value of [A]0, so we need to eliminate [A] from this equation. We can rearrange the integrated rate law equation to solve for t:

t = ln([A]0/[A]) / k

Again substituting the given values, we get:

24.2 s = ln([A]0/[A]) / k

Solving for [A] using the integrated rate law and the value of t from the previous equation:

[A] = [A]0 e^(-kt)

[A] = [A]0 e^(-(8.75 × 10^-12 s^-1)(24.2 s))

[A] = [A]0 e^(-2.118 × 10^-10)

Using this expression for [A], we can substitute into the equation relating [A] and [B], and solve for [A]0:

0.322 M = 2[A]0 - 2[A]

0.322 M = 2[A]0 - 2[A]0 e^(-2.118 × 10^-10)

[A]0 = 0.183 M

Therefore, the initial concentration of A was 0.183 M, which corresponds to answer choice B.

For the First-Order Reaction Below,the Concentration of Product B After