For a first order reaction, rate = $$k[A]$$ . If A₁ and A₂ be the initial and final concentrations of the reactant respectively then,

$${r_1} = k[{A_1}]$$ or, $$0.04 = k[{A_1}]$$ .............. [1]

and $${r_2} = k[{A_2}]$$ or, $$0.03 = k[{A_2}]$$ ....... [2]

$$\therefore$$ $${{[{A_1}]} \over {[{A_2}]}} = {{0.04} \over {0.03}} = {4 \over 3}$$ (Dividing equation [1] by equation [2])

At 10 min, $$10 = {{2.303} \over k}\log {{[{A_0}]} \over {[{A_1}]}}$$, at 20 min, $$20 = {{2.303} \over k}\log {{[{A_0}]} \over {[{A_2}]}}$$

$$\therefore$$ $$20 - 10 = 10 = {{2.303} \over k}\log {{[{A_1}]} \over {[{A_2}]}}$$

or, $$k = {{2.303} \over {10}}\log \left( {{4 \over 3}} \right)$$ or, $$k = 2.855 \times {10^{ - 2}}$$ min^$$-$$1

$$\therefore$$ The half-life for the first order reaction,

$$ = {{0.693} \over k} = {{0.693} \over {2.855 \times {{10}^{ - 2}}}} = 24.27$$ min