Butadiene can undergo the following reaction To form a dimer(two butadiene molecules hooked together). 2c4h8---->c8h12.
The half life for the reaction at a given temperature is 5.92x10-2 sec. The reaction kinetics are second order.
1. If the initial concentration of c4h8 is. 5M, what is the rate constant for the reaction?
2. If the initial concentration of c4h8 is. 1M what will be the concentration of the c4h8 after 3.6x 10 sec
Use the integrated equation and substitute the numbers.
To answer the first question, we need to use the second-order rate law equation, which can be written as:
Rate = k * [C4H8]^2
where Rate is the rate of the reaction and [C4H8] is the concentration of butadiene.
1. To find the rate constant (k) for the reaction, we can rearrange the rate law equation as follows:
Rate / [C4H8]^2 = k
Given that the half-life (t1/2) of the reaction is 5.92 × 10^-2 seconds, we can use the following equation to relate the rate constant to the half-life:
k = 0.693 / t1/2
Plugging in the given half-life value:
k = 0.693 / (5.92 × 10^-2)
k ≈ 11.70 M^-1s^-1
So, the rate constant for the reaction is approximately 11.70 M^-1s^-1.
Now, let's move on to the second question:
2. If the initial concentration of C4H8 is 1M, we can use the integrated rate law equation for a second-order reaction to determine the concentration of C4H8 after a given time (t):
1 / [C4H8]t - 1 / [C4H8]0 = kt
Rearranging the equation, we can solve for [C4H8]t:
[C4H8]t = 1 / (1 / [C4H8]0 + kt)
Given that [C4H8]0 = 1M and t = 3.6 × 10 seconds, we can substitute these values into the equation:
[C4H8]t = 1 / (1 / 1M + (11.70 M^-1s^-1) × (3.6 × 10 seconds))
Calculating this expression:
[C4H8]t ≈ 0.061 M
Therefore, the concentration of C4H8 after 3.6 × 10 seconds will be approximately 0.061 M.