Consider this reaction:→2NH3g+N2g3H2gAt a certain temperature it obeys this rate law.rate =2.89s−1NH3Suppose a vessel contains NH3 at a concentration of 0.640M. Calculate the concentration of NH3 in the vessel 0.380 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.
Question
Consider this reaction:→2NH3g+N2g3H2gAt a certain temperature it obeys this rate law.rate =2.89s−1NH3Suppose a vessel contains NH3 at a concentration of 0.640M. Calculate the concentration of NH3 in the vessel 0.380 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.
Solution
The rate law for the reaction is given as rate = 2.89s^−1[NH3]. This is a first order reaction with respect to NH3.
For first order reactions, we can use the formula:
[NH3]t = [NH3]0 * e^(-kt)
where: [NH3]t is the concentration of NH3 at time t, [NH3]0 is the initial concentration of NH3, k is the rate constant, and t is the time.
Given: [NH3]0 = 0.640 M, k = 2.89 s^-1, and t = 0.380 s.
Substituting these values into the formula, we get:
[NH3]t = 0.640 M * e^(-2.89 s^-1 * 0.380 s)
Calculating the above expression will give us the concentration of NH3 in the vessel 0.380 seconds later.
Let's calculate it:
[NH3]t = 0.640 M * e^(-1.0982)
[NH3]t = 0.640 M * 0.333
[NH3]t = 0.213 M
So, the concentration of NH3 in the vessel 0.380 seconds later is approximately 0.21 M when rounded to two significant digits.
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