Consider the decomposition of N2)5 in carbon tetrachloride (CCl4) at 45 degrees Celsius. 2N2O5 (aq) ↔ 4NO2 (g) + O2 (g) The reaction is first-order in N2O5 with the specific rate constant 6.08 x 10^-4/s. Calculate the reaction rate at these conditions:

a. [N2O5] = 0.200 mol/L
b. [N2O5] = 0.319 mol/L

The first one is supposed to be N2O5

Rate = k[N₂O₅]¹

a. Rate 1 = (6.08 x 10⁻⁴ s⁻¹)(0.200 mol-L⁻¹) = 1.36 x 10⁻⁴ mol/L-s
b. You do ‘b’

To calculate the reaction rate at the given conditions, we will use the rate equation for the decomposition of N2O5:

Rate = k[N2O5]

where k is the specific rate constant.

a. [N2O5] = 0.200 mol/L

To calculate the reaction rate, we need to substitute the given values into the rate equation:

Rate = (6.08 x 10^-4/s) x (0.200 mol/L)
= 1.216 x 10^-4 mol/(L·s)

Therefore, the reaction rate at [N2O5] = 0.200 mol/L is 1.216 x 10^-4 mol/(L·s).

b. [N2O5] = 0.319 mol/L

Similarly, substitute the given values into the rate equation to calculate the reaction rate:

Rate = (6.08 x 10^-4/s) x (0.319 mol/L)
= 1.942 x 10^-4 mol/(L·s)

Therefore, the reaction rate at [N2O5] = 0.319 mol/L is 1.942 x 10^-4 mol/(L·s).

To calculate the reaction rate, we need to use the rate equation for the given reaction, which is first-order in N2O5. The rate equation can be written as follows:

Rate = k[N2O5]

where:
Rate = reaction rate
k = specific rate constant
[N2O5] = concentration of N2O5

a. [N2O5] = 0.200 mol/L

To calculate the reaction rate at this concentration, simply substitute the given concentration into the rate equation.

Rate = k[N2O5]
Rate = (6.08 x 10^-4/s) x (0.200 mol/L)
Rate = 1.216 x 10^-4 mol/(L s)

Therefore, the reaction rate at [N2O5] = 0.200 mol/L is 1.216 x 10^-4 mol/(L s).

b. [N2O5] = 0.319 mol/L

Similarly, to calculate the reaction rate at this concentration, substitute the given concentration into the rate equation.

Rate = k[N2O5]
Rate = (6.08 x 10^-4/s) x (0.319 mol/L)
Rate = 1.942 x 10^-4 mol/(L s)

Therefore, the reaction rate at [N2O5] = 0.319 mol/L is 1.942 x 10^-4 mol/(L s).