In: Chemistry
The gas phase reaction H2 + I2 --> 2HI is second order. Its rate constant at 400 C is 0.0243 dm3/mol s. Calculate delta H, delta S, delta G and the pre-exponential at this temperature. Assume that delta H is constant over this temperature range.
for H2 (g) + I2 (g):::::::::::::::::> 2HI (g)
We have
standard enthalpy of formation...
dHf HI (g) = + 26.48 kJ / mol
dHf H2 (g) = 0
dHf of I2 (g) = + 62.44 kJ/mol
so
our reaction would have
dh = dHf products - reactants
dH = [(2 mol HI) (+ 26.48 kJ / mol)] - [ 0 & 62.44 kJ]
dH = 52.96 kJ - 62.44 kJ
dH reaction = - 9.48 kJ
& We have the standard entropies:
HI ... S = + 206.5 Joules / mol-K
I2 ... S = + 260.6 Joules / mol-K
H2 .. S = + 130.6 Joules / mol-K
so
our reaction would have
dS = S products - reactants
dS = [(2 mol HI) (+ 206.5 Joules / mol-K)] - [ 130.6 &
260.6]
dS = 413 - 391.2
dS = + 21.8 Joules / K
using the formula:
dG = dH - T dS
dG = -9.48 kJ - (298K) (+ 21.8 Joules / K)
dG = -9.48 kJ - (6496 Joules)
dG = -9.48 kJ - (6.50 kiloJoules)
dG = - 15.98 kJ for the reaction that produces 2 moles of HI
which is rounded off to 1 decimal for sig figs as
our answer is
dG reaction = - 16.0 kJ
We know K= Ae^(Ea/RT)
Here Ea for this reaction is not given.According to Standard values Ea for this reaction at 400 C and rate constant 0.0243 is around 180kj.
so A=K/(e^(Ea/RT)).
A=0.0243/e^(-Ea/8.314*673)) ------------------(1)
putting Ea=180kj
we get A=2.5*10^-2 or 0.025.
if any other Ea value is there then substituting in (1) will give
A.