Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
Answer:
a) r = 4.82x10⁻³*[N2O5]
b) 1.16x10⁻⁴ M/s
c) The rate is doubled too (2.32x10⁻⁴ M/s)
d) The rate is halved too (5.78x10⁻⁴ M/s)
Explanation:
a) The rate law of a generic reaction (A + B → C + D) can be expressed by:
r = k*[A]ᵃ*[B]ᵇ
Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).
In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:
r = k*[N2O5]
r = 4.82x10⁻³*[N2O5]
b) Substituing the value of the concentration in the rate law:
r = 4.82x10⁻³*0.0240
r = 1.16x10⁻⁴ M/s
c) When [N2O5] = 0.0480 M,
r = 4.82x10⁻³*0.0480
r = 2.32x10⁻⁴ M/s
So, the rate is doubled too.
d) When [N2O5] = 0.0120 M,
r = 4.82x10⁻³*0.0120
r = 5.78x10⁻⁴ M/s
So, the rate is halved too.
