a first order reaction has rate constant 0.0051 min-1. if we begin with 0.10M cconcentration of the reactant, what concentration of reactant will remain in solution after 3 hours?
THE ANSWER OF THIS QUESTION IS 4.097 *10-3.
KINDLY MATCH THIS ANSWER AND PLEASE WRITE EACH STEP AS I AM NOT GETTING THE STEPS.
Kindly recheck the question as on using the data given in the question, the answer comes out to be 3.994 X 10-2 and not 4.097 X 10-3. Following is the answer as per the given data
Here we will first apply the integrated first order equation which is given as follows
where k is the reaction constant, t is the time in which reaction has occurred, [A]1 is the initial concentration and [A]2 is the concentration after time t. We are given that t = 0.0051 min-1, and t = 3 hours. Therefore we will convert 3 hours to minutes by multiplying with 60. Also, the initial concentration is 0.1M, and we have to calculate the concentration after 180 minutes. So, substituting the values of k, t, [A]1 in the above equation and solving for [A]2, we will get
log ([0.1] / [A]2) = 0.3986 (1)
⇒ -1 - log ([A]2) = 0.3986
or log ([A]2) = - 1.3986
Since we cannot take the antilog of a negative number, so we will add and subtract 1 from the above value. Thus we will now take the antilog of 
. This comes out to be 3.994 X 10-2.
We can also simply take the antilog in equation (1). So we will get
([0.1] / [A]2) = antilog 0.3986 = 2.506
Solving for [A]2, we will get
[A]2 = 3.994 X 10-2.