For a first-order reaction, the time $t$ required for the concentration to change from $a$ to $(a-x)$ is given by the integrated rate law: $$ \begin{aligned} t &= \frac{1}{K} \ln \left( \frac{a}{a-x} \right) \end{aligned} $$ Given: $K = 0.041 \text{ min}^{-1}$ $a = 0.50 \text{ M}$ $a-x = 0.10 \text{ M}$ Substituting the values into the equation: $$ \begin{aligned} t &= \frac{1}{0.041} \ln \left( \frac{0.50}{0.10} \right) \\ &= \frac{1}{0.041} \ln(5) \\ &= \frac{1.6094}{0.041} \\ &= 39.2546 \text{ min} \end{aligned} $$ It will take approximately $39.25$ minutes.