What is the rate of the reaction:
H2SeO3 + 6I- + 4H+ Se + 2I3- + 3H2O
given that the rate law for the reaction at 0°C is
rate = (5.0 × 10^5 L5mol-5s-1)[H2SeO3][I-]^3[H+]^2
The reactant concentrations are [H2SeO3] = 1.5 × 10^-2 M, [I-] = 2.4 × 10^-3 M, [H+] = 1.5 × 10^-3 M.
h k
To find the rate of the reaction, we can use the rate law equation and substitute the given values of reactant concentrations.
Given:
[H2SeO3] = 1.5 × 10^-2 M
[I-] = 2.4 × 10^-3 M
[H+] = 1.5 × 10^-3 M
Rate law equation:
rate = (5.0 × 10^5 L^5mol^-5s^-1) [H2SeO3][I-]^3[H+]^2
Substituting the given concentrations into the rate law equation:
rate = (5.0 × 10^5 L^5mol^-5s^-1) (1.5 × 10^-2 M)(2.4 × 10^-3 M)^3(1.5 × 10^-3 M)^2
To simplify the calculation, let's first calculate the concentration terms:
(1.5 × 10^-2 M)(2.4 × 10^-3 M)^3(1.5 × 10^-3 M)^2
= (1.5 × 10^-2 M)(2.4^3 × 10^-9 M^3)(1.5^2 × 10^-6 M^2)
= (1.5 × 2.4^3 × 1.5^2 × 10^-2 × 10^-9 × 10^-6) M^6
= (1.5 × 13.824 × 2.25 × 10^-2 × 10^-9 × 10^-6) M^6
= (4.73 × 10^-9) M^6
Now, substitute the calculated value back into the rate equation:
rate = (5.0 × 10^5 L^5mol^-5s^-1)(4.73 × 10^-9) M^6
Finally, to find the rate of the reaction, multiply the two values together:
rate = (5.0 × 10^5 L^5mol^-5s^-1)(4.73 × 10^-9) M^6
= 2.365 × 10^-3 L^5mol^-5s^-1
So, the rate of the reaction is 2.365 × 10^-3 L^5mol^-5s^-1.