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Karnataka Board PUCPUC Science 2nd PUC Class 12
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पाठ्यक्रम

First Order Reactions

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Estimated time: 3 minutes
CBSE: Class 12

Key Points: First Order Reactions

A reaction is first order if the rate depends on the first power of concentration of one reactant.

For A → Products:

Time Concentration
t = 0 a
t = t a − x

\[k=\frac{2.303}{t}\log\frac{a}{a-x}\quad\mathrm{or}\quad k=\frac{2.303}{t}\log\frac{[A]_0}{[A]}\]

Also: \[[A]=[A]_0\cdot e^{-kt}\]

Half-life:

\[t_{1/2}=\frac{0.693}{k}\]

  • Half-life is independent of initial concentration — a defining feature of first order reactions.
  • \[t_{75\%}=2\times t_{1/2}\]

Related QuestionsVIEW ALL [35]

The rate constant for the reaction:

\[\ce{2N2O_{(s)} ->2N2O4_{(g)}}\] is 4.98 × 10-4 s-1.

The order of the reaction is ______.

A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.

(Given : log = 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)

Slove: \[\ce{2NOBr -> 2NO_{2(g)} + Br_{2(g)}}\]

For the above reaction, the rate law is rate = k[NOBr]2. If the rate of reaction is 6.5 × 10−6 mol L−1 s−1 at 2 × 10−3 mol L−1 concentration of NOBr, calculate the rate constant k for the reaction.

Straight line graph for first order reaction is obtained between ____________.

 A first order reaction is 50% complete in 25 minutes. Calculate the time for 80% completion of the reaction.

Following data are obtained for reaction :

N2O5 → 2NO2 + 1/2O2

t/s 0 300 600
[N2O5]/mol L–1 1.6 × 10-2 0.8 × 10–2 0.4 × 10–2

1) Show that it follows first order reaction.

2) Calculate the half-life.

(Given log 2 = 0.3010, log 4 = 0.6021)

How will you represent first order reactions graphically?

Show that `t_(1/2)= 0.693/k` for first reaction.

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