The gas-phase reaction between hydrogen and iodine H 2 (g) + I 2(g) <--> 2HI (g)

Question

The gas-phase reaction between hydrogen and iodine

H2(g) + I2(g) <--> 2HI(g)

proceed with a rate constant for the forward reaction at 700 deg C of 138 L/mol*s and an activation energy of 165kj/mol.

A) Calculate the activation energy of the reverse reaction given that the standard enthalpy of formation for HI(g) is 26.48kJ/mol and for I2(g) is 62.44kJ/mol.

B)Calculate the rate constant for the reverse reaction at 700 deg C. (assume A in the equation k= Ae(-EA/RT) is the same for both forward and reverse reactions.)

C) Calculate the rate of the reverse reaction if the concentration of HI is 0.200M. The reverse reaction is second order in HI.

Please show work

Explanation / Answer

A) dH of forward reaction = 2 dH HI - dH I2 = 2( 26.48) -62.44 = -9.48 KJ/mol

this means products are lower in energy compared to reactants

hence activation energy for reverse = 9.48 + 165 = 174.48 KJ/mol

B) k rerverse / k forward = exp ( - E (f) /RT) / exp ( - E(b) / RT)

k reverse / 138 = exp ( - 174480/8.314 x 973) / exp ( - 165000/8.314x973)

k reverse = 138 x ( 4.294 x 10^-10) / ( 1.386 x 10^-9)

= 42.75 L/mols

C) rate = k [HI]^2

= 42.75 x ( 0.2^2)

= 1.71 mol/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.