Two substances X and Y decompose via first-order kinetics. The half-life of X is 20 min and that of Y is 40 min . Initially, the rate of decomposition of X is exactly equal to the rate of decomposition of Y. The time it will take for the concentration of Y to become 4 times the concentration of X is ________ min.

Step-by-step solution

For a first-order reaction, the rate of decomposition is given by Rate = k[A] . Given that the initial rates are equal: k_X[X]₀ = k_Y[Y]₀ Using the relation k = 2 t_ 1/2 , we get: ( 2 20 )[X]₀ = ( 2 40 )[Y]₀ [X]₀ 20 = [Y]₀ 40 [Y]₀ = 2[X]₀ We need to find the time t when [Y]_t = 4[X]_t . Using the integrated rate law [A]_t = [A]₀ ( 1 2 )^ t t_ 1/2 : [Y]₀ ( 1 2 )^ t 40 = 4[X]₀ ( 1 2 )^ t 20 Substitute [Y]₀ = 2[X]₀ : 2[X]₀ ( 1 2 )^ t 40 = 4[X]₀ ( 1 2 )^ t 20 Divide both sides by 2[X]₀ : ( 1 2 )^ t 40 = 2 ( 1 2 )^ t 20