Question

It is often stated in many introductory chemistry texts that near room temperature, a reaction rate doubles if the temperature increases by 10°. Calculate the activation energy of a reaction that obeys this rule exactly. Would you expect to find this rule violated frequently? [R=1.987 cal/mol•K=8.314 J/mol•K]?

2) An endothermic reaction A → B has a positive internal energy change, ΔE, or enthalpy change,

ΔH. In such a case, what is the minimum value that the activation energy can have?

3) Assume that the reaction 5Br- + BrO3- + 6H+ → 3Br2 + 3H2O goes by the mechanism

k1

(1) BrO3- + 2H+ → H2BrO3+ (fast reaction)

k-1

(2) H2BrO3+ → BrO3- + 2H+ (fast reaction)

k2

(3) Br- + H2BrO3+ → Br2O2 + H2O (slow reaction)

k3

(4) Br2O2 + 4H+ + 4Br- → 3Br2 + 2H2O (fast reaction)

Deduce the rate law that is consistent with this mechanism.

(b) Express the rate constant for the overall reaction in terms of rate constants for the individual

steps. {Hint: The rate law depends on the concentrations of [H+], [Br-], and [BrO3-].}

Answer 1

1) let the activation energy be H.so,

room temp. = 298 K

so,

ln(K1/k2) = (H/R)*(1/T2 - 1/T1)

or ln(2) = (H/8.31)*(1/298 - 1/308)

or H=52.868 kJ/mol

2)since dH=dE+w

now since both dE and dH are positive, the minimum value of activation energy will be (dH-dE).

3)K=[Br2O2][H2O}/[Br-][H2BrO3+]

K1=[H2BrO3+]/[H+]^2[BrO3-]

K2 = [H+]^2[BrO3-]/[H2BrO3+]

K4 = [H2O]^2([Br2]^3 /([Br-]^4 *[H+]]^4*[Br2O2])

substitutuing the values of Br-, H2BrO3+ and Br2O2 from K1, K2 and K3 into K, we get the desired rate law.