Question

In a first order reaction t ½ = 1000 minutes calculate its rate constant​

Answer 1

Given : first order reaction t ½ = 1000 mins

Half time = 1000 mins

To Find : rate constant of first order reaction

Solution:

first order reaction

$A=A_0e^{-kt}$

=> t = (1/k) ln(A₀/A)

A₀ = initial

A = remaining

t = time

k =  rate constant  

t will be half time if  A = A₀/2

=> A₀/A = 2

Half time = 1000 mins

=> 1000 = (1/ k ) ln(2)

=> k = ln(2)/1000

=> k = 6.93 * 10⁻⁴   per minute

or k = 1.155 * 10⁻⁵   per Sec

Hence rate constant​  6.93 * 10⁻⁴   per minute  or 1.155 * 10⁻⁵   per Sec

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Answer 2

• For school exams, please follow the above method by @Amitnrw sir.

• But for MCQs, follow this method, it's very fast !

For a 1st order reaction:

$t_{ \frac{1}{2} } = \dfrac{0.693}{k}$

• "k" is the rate constant for the reaction kinetics.

$\implies \: 1000 = \dfrac{0.693}{k}$

$\implies \: k= \dfrac{0.693}{1000}$

$\implies \: k= 6.93 \times {10}^{ - 4} \: {min}^{ - 1}$

So, rate constant is 6.93 × 10^(-4) per min.