Question

The rate of firrst order reaction is 0.04 mole at 10 min amd 0.03 mole at 20 min after initiation what is tge half life of the rnx

Answer 1

Reaction Type: 1st order reaction.

Formula used: $\tt{k = \bigg(\dfrac{2.303}{t_{2}-t_{1}}\bigg) \log \bigg(\dfrac{[R_{1}]}{[R_{2}]}\bigg)}$

Given:

$\tt{T_{1}=10\;min\;\;\&\;\;T_{2}=20\;min}$

Therefore,

$\tt{T_{2} - T_{1} = 10\;min}$

[R1] = 0.04 mol/L

[R2] = 0.03 mol/L

Substituting the values in the above equation

$\tt{k = \bigg(\dfrac{2.303}{t_{2}-t_{1}}\bigg) \log \bigg(\dfrac{[R_{1}]}{[R_{2}]}\bigg)}$

$\tt{k=\bigg(\dfrac{2.303}{10}\bigg) \log \bigg(\dfrac{0.04}{0.03}\bigg)}$

$\tt{k=0.2303\;[\log(0.04)-\log(0.03)]}$

$\tt{k=0.2303\;[-1.3979-(-1.5228)]}$

$\tt{k=0.02876\;min}$

For the first order reaction half life $\tt{t_{1/2}}$ is given by

$\tt{t_{1/2}=\dfrac{0.693}{k}}$

We know, k = 0.02876 we get

$\tt{t_{1/2}=\dfrac{0.693}{0.02876}}$

$\tt{t_{1/2}=24.096\;min}$

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