What is the activation energy for the forward and reverse reactions?

Question

At 650K, rate constant for forward reaction HI(g)=1/2H2 (g)+ 1/2 I2(g) is k=2.15 x 10^-8 L/mol sec.

At 700K, rate constant is 2.39 x 10^-7 1/Ms

the delta H for reverse is 5.12 kJ/mol for reverse rxn.

Answer 1

Use Arrhenius equation to determine activation energy of forward reaction:

k₊(T₁) = A₊·exp{- Ea₊/(R·T₁)}

k₊(T₂) = A₊·exp{- Ea₊/(R·T₂)}

=>

k₊(T₂)/k₊(T₁) = A·exp{- Ea₊/(R·T₂)} / A·exp{- Ea/(R·T₁}

<=>

k₊(T₂)/k₊(T₁) = exp{ (Ea₊/R)·(1/T₁ - 1/T₂) }

<=>

ln(k₊(T₂)/k₊(T₁)) = (Ea₊/R)·(1/T₁ - 1/T₂)

=>

Ea₊ = R · ln(k₊(T₂)/k₊(T₁)) / (1/T₁ - 1/T₂)

= 8.314472Jmol⁻¹K⁻¹ · ln(2.39×10⁻⁷Lmol⁻¹s⁻¹ / 2.15×10⁻⁸Lmol⁻¹s⁻¹) / ( 1/650K - 1/750K)

= 182.22 kJ/mol

The difference of the activation energy is equal to th enthalpy of reaction.

You can deduce this by considering the equilibrium constant of the reaction. By kinetic definition it is defined as the ratio of forward to reverse rate constant:

K = k₊ / k₋

= A₊·exp{- Ea₊/(R·T)} / A₋·exp{- Ea₋/(R·T)}

= (A₊/A₋) ·exp{- (Ea₊ - Ea₋) /(R·T)}

Van't Hoff equation, which is based on general thermodynamic considerations, relates equilibrium constant to temperature as:

K = K₀ ·exp{- ΔH° /(R·T)}

Since both equation need to be consistent, the argument of the exponential function must be the same. Hence:

Ea₊ - Ea₋ = ΔH°

=>

Ea₋ = Ea₊ - ΔH°

= 182.22 kJ/mol - (-5.12kJ/mol) =  187.34kJ/mol

Note:

ΔH° is the relations above , ΔH° of forward reaction. It has the same magnitude as for reverse reaction but different sign.