The reaction described by this equation
O3(g) + NO(g) -- O2(g) + NO2(g)
has the following rate law at 310 K.
rate of reaction = 3.0x10^6 M^-1 s^-1)[O3][NO]
Given that [O3]=2.0× 10–4 M and [NO]=2.0× 10–5 M at t=0, calculate the rate of the reaction at t=0.
.012
rate rxn = k[O3][NO]
Substitute for k and each of the cncns (given in the problem) and solve for rate of rxn.
To calculate the rate of the reaction at t=0, we need to substitute the initial values of [O3] and [NO] into the rate law equation.
Given:
[O3] = 2.0 × 10^–4 M
[NO] = 2.0 × 10^–5 M
rate of reaction = (3.0 × 10^6 M^–1 s^–1) [O3] [NO]
Substituting the values:
rate of reaction = (3.0 × 10^6 M^–1 s^–1) (2.0 × 10^–4 M) (2.0 × 10^–5 M)
Let's simplify this calculation step by step:
Step 1: Simplify the scientific notation for [O3] and [NO].
[O3] = 2.0 × 10^–4 M = 2.0 × 10^–4 M
[NO] = 2.0 × 10^–5 M = 2.0 × 10^–5 M
Step 2: Multiply the values of [O3] and [NO] together.
[O3] [NO] = (2.0 × 10^–4 M) (2.0 × 10^–5 M) = 4.0 × 10^–9 M^2
Step 3: Multiply the product of [O3] and [NO] by the rate constant.
rate of reaction = (3.0 × 10^6 M^–1 s^–1) (4.0 × 10^–9 M^2)
Step 4: Simplify the units.
rate of reaction = (3.0 × 10^6) (4.0 × 10^–9) M^–1 M^2 s^–1
Step 5: Combine the exponents of M.
rate of reaction = 12.0 × 10^–9 M s^–1
Step 6: Convert the rate of reaction to scientific notation.
rate of reaction = 1.2 × 10^–8 M s^–1
Therefore, the rate of the reaction at t=0 is 1.2 × 10^–8 M s^–1.