The rate constant for a First order reaction at 560 ~K is 1.5 10⁻⁶ per second. If the reaction is allowed to take place for 20 hours, what percentage of the initial concentration would have converted to products?

Step-by-step solution

For a first order reaction, the integrated rate equation is given by k = 2.303 t ( [A]₀ [A]_t ) . Given values are k = 1.5 10⁻⁶ s ⁻¹ and t = 20 hours = 20 3600 s = 72000 s . Substituting these values into the equation: 1.5 10⁻⁶ = 2.303 72000 ( [A]₀ [A]_t ) ( [A]₀ [A]_t ) = 1.5 10⁻⁶ 72000 2.303 = 0.108 2.303 0.046895 . Taking the antilog: [A]₀ [A]_t = 10^ 0.046895 1.11397 . The fraction of reactant remaining is [A]_t [A]₀ = 1 1.11397 0.89766 . The fraction converted to products is 1 - [A]_t [A]₀ = 1 - 0.89766 = 0.10