# 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.

• 1

k=0.0051

Ro=0.10M ; R=?

t=3hr=180min

k=2.303/t  logRo/R

using above eq calulate R

• -10

i have used the same equation but the problem is that we need to take antilog at last but the value is negative so i don't know how to take antilog of negative value.

• -10

No,after calculation we get;; log0.10/R=0.3986

For ANTILOG of 0.3986 just look the tables,like in first look for 0.39 then 8 in the same row and add mean difference of 6!!

• -17
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