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}.

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