The rate constant for the reaction below is 6.2 x 10−5 mol l−1 s −1. if the initial concentration of a is 0.0500 m, what is its concentration after 115 s?

The correct answer is 0.04287 M

It is given that the rate constant of the reaction is 6.2 × 10⁻⁵ mol/L/s

Thus, as the rate of the reaction is in mol/L/s, it is a zero order reaction.  

The initial concentration of a is 0.0500 M, and there is a need to find the concentration of a after 115 seconds. This can be done with the help of the equation:  

[A]o -[A] = kt

[A] = [A]o - kt

= (0.0500 M) - (6.2 × 10⁻⁵) (115 s)

= 0.04287 M

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

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