A compound X breaks down on a metal surface…

Step 1. The slope. The concentration falls from $0.620\ \text{mol/L}$ to $0.314\ \text{mol/L}$ in $20\ \text{min}$. The fall is $0.306\ \text{mol/L}$, so $k = \dfrac{0.306\ \text{mol/L}}{20\ \text{min}} = 0.0153\ \text{mol L}^{-1}\,\text{min}^{-1}$.

Step 2. Half-life. $t_{1/2} = \dfrac{0.620\ \text{mol/L}}{2 \times 0.0153\ \text{mol L}^{-1}\,\text{min}^{-1}} = 20.3\ \text{min}$.

Step 3. All used up. $t = \dfrac{0.620\ \text{mol/L}}{0.0153\ \text{mol L}^{-1}\,\text{min}^{-1}} = 40.5\ \text{min}$.