Step 1. Here V (volume of standard $KMnO_4$ needed to titrate the sample) is directly proportional to the REMAINING concentration of $H_2O_2$ at each time (unlike Question 28, where the gas evolved tracked the amount reacted) -- so the first order formula applies directly as $k=\dfrac{2.303}{t}\log\dfrac{V_0}{V_t}$.

Step 2. At $t=10$: $k=\dfrac{2.303}{10}\log\dfrac{46.1}{29.8}=0.2303\times\log(1.547)=0.2303\times0.1895=0.04364\ \text{min}^{-1}$.

Step 3. At $t=20$: $k=\dfrac{2.303}{20}\log\dfrac{46.1}{19.3}=0.11515\times\log(2.389)=0.11515\times0.3782=0.04356\ \text{min}^{-1}$. …

Recognise that the titration volume V is directly proportional to the remaining $H_2O_2$ concentration, compute k at each interval …

• Treating V here the same way as the gas-volume V in Question 28 (i.e. using $V_0-V_t$ instead of $V_t$ directly) -- the two problems have OPPOSITE physical setups: Q28's gas volume …