# Formula for rate constant for the first order reaction

My lecturer mentioned that the formula for the rate constant $$k$$ for the first order reaction is

$$k = \frac{2.0303}{t_{1/2}}\,\log\frac{[\ce{A}]_0}{[\ce{A}]_{t_{1/2}} - [\ce{A}]_0},\tag{1}$$

where $$t_{1/2}$$ is the half-life; $$[\ce{A}]_0$$ is the initial concentration; $$[\ce{A}]_{t_{1/2}}$$ is the concentration at half-life.

In which what I know recently is that the rate constant for the first order reaction is

\begin{align} & & \ln\frac{[\ce{A}]_t}{[\ce{A}]_0} &= -kt \tag{2}\\ &\implies & [\ce{A}]_t &= [\ce{A}]_0\,\mathrm e^{-kt}\tag{3} \end{align}

I ask my teacher, but he said I should go and do more research, but I couldn't find it.

• Your first equation has some typos, the second and third are correct. You need to use $[A]_{1/2}=[A_0]/2$ and let $t\to t_{1/2}$ Feb 26 '21 at 8:49
• Indeed, there are typos!
– PV.
Feb 26 '21 at 14:45
• Can anyone help me with the derivation of the first one @porphyrin Feb 26 '21 at 15:25

I am attaching a written answer as typing the log and other exponents is hard :(

The two equations are pretty the same. The number 2.30 is used to convert the natural logarithm in decimal logarithm: $$\ln 10 = 2.303$$. Also, substitute $$[A]_{1/2} = [A]_0 /2$$, from the definition. The concentration at the half-life is exactly the half of the initial concentration, then you will obtain $$\ln 2$$ inside the logarithm term.

• Except for the typo that makes the argument of the logarithm in equation (1) of the OP negative. Feb 26 '21 at 17:19

The half-life $$t_½$$ is the time it takes for the concentration of the first order reaction to decrease by a factor of two. If we plug this into equation (2) given by the OP, we get

$$\ln\frac{1}{2} = -k t_½$$

We solve this for $$k$$ to get:

$$k = \frac{ \ln(2) }{t_½} \approx \frac{0.693}{t_½}$$

My lecturer mentioned that the formula for the rate constant $$k$$ for the first order reaction is $$k = \frac{2.303}{t_{1/2}}\,\log\frac{[\ce{A}]_0}{[\ce{A}]_{t_{1/2}} - [\ce{A}]_0},\tag{1}$$

The argument of the logarithm is two (once you swap the two parts of the difference shown in the denominator - probably a typo), and $$2.303 \log(x)$$ is the same as $$\ln(x)$$, so this is correct but probably just a step on the way to the relationship given above that is more direct and useful. If you do the math, you will find

$$2.303 \log(2) \approx 0.693$$

• @porphyrin - I fixed it, thanks (another "typo" in the formula reported by the OP). Feb 27 '21 at 11:36

I think what OP need is the derivation of the equation $$(1)$$ in the question (as indicated in the OP's comment, "an anyone help me with the derivation of the first one").

The first part of this derivation, the formula for the rate constant $$k$$ for the first order reaction, is done in the last part of Chemistry student's answer, and hence I'd not repeat it. Thus,

$$[\ce{A}]_t = [\ce{A}]_0\,\mathrm e^{-kt}\tag{3}$$ Or when the rearrangement of the equation $$(3)$$ gives: $$-kt = \ln\frac{[\ce{A}]_t}{[\ce{A}]_0}$$ $$\implies kt = \ln\frac{[\ce{A}]_0 }{[\ce{A}]_t} \tag{2}$$

In equations $$(2) \ \& \ (3)$$, $$[\ce{A}]_0$$ is the initial concentration and $$[\ce{A}]_{t}$$ is the concentration at time $$t$$. If $$t_{1/2}$$ is the half-life and $$[\ce{A}]_{t_{1/2}}$$ is the concentration at half-life, we can rewrite the equation $$(2)$$ as: $$kt_{1/2} = \ln\frac{[\ce{A}]_0 }{[\ce{A}]_{t_{1/2}}} \tag{4}$$

Actually, $$[\ce{A}]_{t_{1/2}}$$ is $$\frac{1}{2}[\ce{A}]_0$$, and therefore, $$[\ce{A}]_{t_{1/2}}$$ can be written as $$[\ce{A}]_0 - [\ce{A}]_{t_{1/2}}$$. Substituting $$[\ce{A}]_{t_{1/2}} = [\ce{A}]_0 - [\ce{A}]_{t_{1/2}}$$ in the equation $$(4)$$, you get:

$$kt_{1/2} = \ln\frac{[\ce{A}]_0 }{[\ce{A}]_0 - [\ce{A}]_{t_{1/2}}} \tag{5}$$

Applying $$\ln x = 2.303 \log x$$ in the equation $$(5)$$ and rearranging it to get $$k$$, we finally get:

$$k = \frac{2.303}{t_{1/2}} \log \frac{[\ce{A}]_0 }{[\ce{A}]_0 - [\ce{A}]_{t_{1/2}}} \tag{6}$$

As evident and shown by others, OP's lecturer mentioned formula has two errors: one is making the $$\log$$ function negative, and the other is converting $$\ln x$$ to $$\log 2.0303 x$$, which is incorrect:

$$\ln x = \log_e x = \frac{\log_{10} x}{\log_{10} e} = \frac{\log_{10} x}{\frac{\log_{e} e}{{\log_{e} 10} }} = (\log_{10} x)(\log_{e} 10) = 2.303\log_{10} x$$

Thus, the equation $$(6)$$ is the most accurate equation in that regards.