The rate constant for the thermal decomposition at 85 degrees Celsius of dinitrogen pentoxide equals 6.2 x 10 raised to -2 min. As determined by experiment this decomposition is first order--
2 N2O5 ----> 4 NO2 + O2
Calculate the rate of the reaction when N2O5= .40 M
To calculate the rate of the reaction, we can use the first-order rate equation:
Rate = k[A]
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] is the concentration of reactant A
In this case, the reactant A is N2O5, and its concentration is given as 0.40 M. We also know the value of the rate constant, which is 6.2 x 10^(-2) min^(-1).
Substituting the given values into the equation, we can calculate the rate of the reaction:
Rate = (6.2 x 10^(-2) min^(-1))(0.40 M)
Rate = 2.48 x 10^(-2) M/min
Therefore, the rate of the reaction when [N2O5] = 0.40 M is 2.48 x 10^(-2) M/min.
To calculate the rate of the reaction, we need to use the first-order rate equation:
Rate = k * [N2O5]
Given:
Rate constant (k) = 6.2 x 10^-2 min^-1
[N2O5] = 0.40 M
Plugging these values into the rate equation:
Rate = (6.2 x 10^-2 min^-1) * (0.40 M)
= 2.48 x 10^-2 M/min
Therefore, the rate of the reaction when [N2O5] = 0.40 M is 2.48 x 10^-2 M/min.