In aqueous solution, hypobromite ion, BrO-, reacts to produce bromate ion, BrO3 -, and bromide ion, Br-, according to the following chemical equation.
3BrO-(ag) → BrO3 -(ag)+2Br-(ag)
A plot of 1/[BrO-] vs. time is linear and the slope is equal to 0.056 M-1s-1. If the initial concentration of BrO- is 0.80 M, how long will it take one-half of the BrO- ion to react?
*Show steps please* thanks!
This is second order because the k constant has different units for each order, so you'll just have to remember them?
The half-life equation for 2nd order is 1/(k[Ao]) = t1/2
k=0.056 M-1s-1
Ao= 0.80M
1/(0.056x0.80) = 22.3s
I believe the reaction is second order.
Why did the hypobromite ion go to therapy? Because it was feeling bromo-dide!
To determine how long it will take for half of the BrO- ions to react, we can use the integrated rate law for a first-order reaction.
The integrated rate law equation is: ln([BrO-]t/[BrO-]0) = -kt
Where:
[BrO-]t is the concentration of BrO- at time t
[BrO-]0 is the initial concentration of BrO- (0.80 M in this case)
k is the rate constant (0.056 M-1s-1 in this case)
t is the time
We can rearrange the equation to solve for t:
ln([BrO-]t/[BrO-]0) = -kt
[BrO-]t/[BrO-]0 = e^(-kt)
[BrO-]t = [BrO-]0 * e^(-kt)
Since we're interested in the time it takes for half of the BrO- ions to react, we can set [BrO-]t equal to half of [BrO-]0:
[BrO-]0 * e^(-kt) = 0.5 * [BrO-]0
e^(-kt) = 0.5
Taking the natural logarithm of both sides:
ln(e^(-kt)) = ln(0.5)
-kt = ln(0.5)
Now we can solve for t:
t = -ln(0.5) / k
Plugging in the given value of k (0.056 M-1s-1), we can calculate t:
t = -ln(0.5) / 0.056 M-1s-1
t ≈ 12.43 s
So, it will take approximately 12.43 seconds for half of the BrO- ions to react. Keep an eye on those bromine ions, they can be quite reactive!
To determine how long it will take for one-half of the BrO- ion to react, we can use the integrated rate law for a first-order reaction. In this case, the reaction is first-order with respect to BrO-.
The integrated rate law for a first-order reaction is given by:
ln([A]t/[A]0) = -kt
where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.
In this case, we are given that the slope of the graph is equal to 0.056 M-1s-1. The slope of the graph is equal to -k, so we can calculate the rate constant:
k = -slope = -0.056 M-1s-1
Now, we need to determine the time it takes for one-half of the BrO- ion to react. One-half of the initial concentration of BrO- is 0.5 * 0.80 M = 0.40 M.
Using the integrated rate law, we can rearrange the equation to solve for time:
ln([A]t/[A]0) = -kt
ln([BrO-]t/0.80 M) = -0.056 M-1s-1 * t
Substituting [BrO-]t = 0.40 M and solving for t, we can calculate the time it takes for half of the BrO- ion to react:
ln(0.40 M/0.80 M) = -0.056 M-1s-1 * t
ln(0.50) = -0.056 M-1s-1 * t
Using the natural logarithm (ln) function on both sides:
-0.693 = -0.056 M-1s-1 * t
Solving for t:
t = -0.693 / (-0.056 M-1s-1)
t ≈ 12.38 s
Therefore, it will take approximately 12.38 seconds for one-half of the BrO- ion to react.
The graph tells you it is a first order reaction and the slope of the line = 0.056 = k
Then k = 0.693/t1/2 and solve for t1/2