The second‐order decomposition of HI has a rate constant of 1.80 × 10-3 L mol-1 s-1. How muchHI remains after 27.3 s if the initial concentration of HI is 4.78 mol L-1?
Question
The second‐order decomposition of HI has a rate constant of 1.80 × 10-3 L mol-1 s-1. How muchHI remains after 27.3 s if the initial concentration of HI is 4.78 mol L-1?
Solution
The reaction is a second-order reaction, which means the rate of the reaction is proportional to the square of the concentration of the reactant. The formula for the concentration of a reactant in a second-order reaction at time t is:
1/[A]t = kt + 1/[A]0
where: [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is the time.
Given: [A]0 = 4.78 mol L-1, k = 1.80 × 10-3 L mol-1 s-1, and t = 27.3 s.
We can substitute these values into the formula:
1/[A]t = (1.80 × 10-3 L mol-1 s-1)(27.3 s) + 1/(4.78 mol L-1)
Solving this equation will give us the concentration of HI at time t = 27.3 s.
1/[A]t = 0.04914 L mol-1 + 0.2092 L mol-1
1/[A]t = 0.25834 L mol-1
To find [A]t, we take the reciprocal of both sides:
[A]t = 1 / 0.25834 L mol-1
[A]t = 3.87 mol L-1
So, after 27.3 seconds, the concentration of HI is 3.87 mol L-1.
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