Question

The decomposition of NO2 is a 1st order reaction. At room temperature it takes 32mins for 0.05M NO2 to decrease to 0.025M. How long will it take 0.025M NO2 to decrease to .0.0125M?

Answer 1

Answer on Question #57283 – Chemistry – Inorganic Chemistry

Question

The decomposition of $\mathrm{NO}_2$ is a 1st order reaction. At room temperature it takes 32 mins for $0.05\mathrm{M}\ \mathrm{NO}_2$ to decrease to 0.025M. How long will it take $0.025\mathrm{M}\ \mathrm{NO}_2$ to decrease to 0.0125M?

Solution

Speed of reaction of the first order is described by the equation

$$w = - \frac {d C _ {A}}{d t} = k _ {1} \cdot C _ {A}. \tag {1}$$

We will divide variables in the equation (1)

$$- \frac {d C _ {A}}{C _ {A}} = k _ {1} d t$$

we will integrate the last equation

$$\int_ {C _ {0}} ^ {C} \left(- \frac {d C _ {A}}{C _ {A}}\right) = \int_ {C _ {0}} ^ {C} - d \ln C _ {A} = \int_ {t = 0} ^ {t} k _ {1} d t,$$

where $C_0$ is initial concentration of substance A in an initial timepoint of $t = 0$.

The resultant equation has an appearance:

$$\ln C = \ln C _ {0} - k _ {1} t, \tag {2}$$

or

$$\ln \left(\frac {C _ {0}}{C}\right) = k _ {1} t$$

From here, $C = C_0 \cdot e^{-k_1 t}$

Semi-transformation time for reaction of the first order doesn't depend on initial concentration of substance

$$t _ {1 / 2} = \frac {1}{k _ {1}} \ln \frac {C _ {0}}{0 . 5 \cdot C _ {0}} = \frac {\ln 2}{k _ {1}}$$

0,025M is a half from 0,05M. and 0,0125M is a half from 0,025M

Answer: 32 mins

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