The reaction, N2O5 = 2NO2+1/2O2 is first order in N2O5 with rate constant 6.2 x 10^-4 s^-1. What is the value of rate of reaction when [N2O5] = 1.25 mol L^-1 s^-1 ?
rate = k(N2O5)^1
You know k and (N2O5) solve for rate. Note: I believe you meant (N2O5) = mols/L^-1.
To determine the rate of the reaction, we can use the rate law expression:
rate = k[N2O5]^m
where:
k = rate constant
[N2O5] = concentration of N2O5
m = order of the reaction with respect to N2O5
In the given reaction, N2O5 is first order, so m = 1.
Plugging in the given values:
k = 6.2 x 10^-4 s^-1
[N2O5] = 1.25 mol L^-1
Using the rate law expression, we have:
rate = k[N2O5]^m
= (6.2 x 10^-4 s^-1)(1.25 mol L^-1)^1
= 7.75 x 10^-4 mol L^-1 s^-1
Therefore, the rate of the reaction when [N2O5] = 1.25 mol L^-1 s^-1 is 7.75 x 10^-4 mol L^-1 s^-1.
To determine the rate of reaction when [N2O5] = 1.25 mol L^-1 s^-1, we need to use the rate equation for the reaction:
rate = k * [N2O5]^n
In this case, the reaction is first-order with respect to N2O5, so n = 1. The rate constant, k, is given as 6.2 x 10^-4 s^-1.
Now we can substitute the given values into the rate equation:
rate = (6.2 x 10^-4 s^-1) * (1.25 mol L^-1)^1
Simplifying the expression, we have:
rate = (6.2 x 10^-4 s^-1) * (1.25 mol L^-1)
Let's calculate the numerical value:
rate = 7.75 x 10^-4 mol L^-1 s^-1
Therefore, the value of the rate of reaction when [N2O5] = 1.25 mol L^-1 is 7.75 x 10^-4 mol L^-1 s^-1.