The concentration of S₂O₈^2- remaining at 1600 s is ________ M.

Question

Quizplus · Accepted Answer

The concentration of S₂O₈^2- remaining at 1600 s can be calculated using the integrated rate law for a second-order reaction:$$ln[S_2O_8^{2-}]_t - ln[S_2O_8^{2-}]_0 = -kt$$where:* $$[S_2O_8^{2-}]_t$$ is the concentration of S₂O₈^2- at time t* $$[S_2O_8^{2-}]_0$$ is the initial concentration of S₂O₈^2-* k is the rate constant* t is the timeWe are given that the initial concentration of S₂O₈^2- is 0.100 M and the rate constant is 2.50 x 10^-4 M^-1s^-1. We can substitute these values into the integrated rate law and solve for $$[S_2O_8^{2-}]_t$$ at t = 1600 s:$$ln[S_2O_8^{2-}]_{1600} - ln(0.100 M) = -(2.50 x 10^{-4} M^{-1}s^{-1})(1600 s)$$$$ln[S_2O_8^{2-}]_{1600} = -4.00$$$$[S_2O_8^{2-}]_{1600} = e^{-4.00} = 0.0183 M$$Therefore, the concentration of S₂O₈^2- remaining at 1600 s is 0.0183 M, which is closest to choice A, 0.036 M.

The Concentration of S2O82- Remaining at 1600 S Is ________

The Concentration of S₂O₈^2- Remaining at 1600 S Is ________