At a certain temperature this reaction follows second-order kinetics with a rate constant of ·0.0577M−1s−1:→NH4OHaq+NH3aqH2OaqSuppose a vessel contains NH4OH at a concentration of 1.16M. Calculate the concentration of NH4OH in the vessel 53.0 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.M
Question
At a certain temperature this reaction follows second-order kinetics with a rate constant of ·0.0577M−1s−1:→NH4OHaq+NH3aqH2OaqSuppose a vessel contains NH4OH at a concentration of 1.16M. Calculate the concentration of NH4OH in the vessel 53.0 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.M
Solution 1
The reaction follows second-order kinetics, which means the rate of the reaction is proportional to the square of the concentration of the reactant. The rate equation for a second-order reaction 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 = 1.16 M, k = 0.0577 M^-1s^-1, and t = 53.0 s.
Substitute these values into the rate equation:
1/[A]t = (0.0577 M^-1s^-1 * 53.0 s) + 1/1.16 M 1/[A]t = 3.0581 M^-1 + 0.8621 M^-1 1/[A]t = 3.9202 M^-1
Solving for [A]t gives:
[A]t = 1 / 3.9202 M^-1 [A]t = 0.26 M
So, the concentration of NH4OH in the vessel 53.0 seconds later is 0.26 M.
Solution 2
The given reaction follows second-order kinetics, which means the rate of the reaction is proportional to the square of the concentration of the reactant. The rate equation for a second-order reaction is given by:
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 = 1.16 M, k = 0.0577 M^-1s^-1, and t = 53.0 s.
Substituting these values into the rate equation:
1/[A]t = (0.0577 M^-1s^-1 * 53.0 s) + 1/1.16 M 1/[A]t = 3.0581 M^-1 + 0.8621 M^-1 1/[A]t = 3.9202 M^-1
Therefore, the concentration of NH4OH in the vessel 53.0 seconds later, [A]t, is:
[A]t = 1/3.9202 M^-1 [A]t = 0.26 M
So, the concentration of NH4OH in the vessel 53.0 seconds later is approximately 0.26 M, rounded to two significant digits.
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