In: Chemistry
Rate constants for the gas-phase decomposition of hydrogen iodide, 2 HI(g) → H2 (g) + I2 (g), are listed in the following table:
Temperature (Celcius) k(M-1s-1)
283 3.52*10-7
356 3.02*10-5
393 2.19*10-4
427 1.16*10-3
508 3.95*10-2
(a) Find the activation energy (in kJ/mol) using all five data points. (b) Calculate Ea from the rate constants at 283 °C and 508 °C. (c) Given the rate constant at 283 °C and the value of Ea obtained in part (b), what is the rate constant at 293 °C?
(a) The given reaction is
2 HI(g) → H2 (g) + I2 (g)
The values are tabulated below
T(°C) | T(K) | 1/T(K-1) | k(M-1s-1) | lnk |
283 | 556 | 0.001799 | 3.52E-07 | -14.8596 |
356 | 629 | 0.00159 | 3.02E-05 | -10.4077 |
393 | 666 | 0.001502 | 2.19E-04 | -8.42644 |
427 | 700 | 0.001429 | 1.16E-03 | -6.75934 |
508 | 781 | 0.00128 | 3.95E-02 | -3.23145 |
The plot is shown below
The Arrhenius equation is
lnk=(-Ea/R)(1/T)+lnA, where k=rate constant, Ea=activation energy, R=gas constant=8.314 J/mol.K, A=pre exponential factor.
The plot lnk vs (1/T) gives a straight line with slope=-Ea/R and y-intercept=lnA.
from the above plot, slope=-Ea/R=-22398.011 K
Therefore activation energy, Ea=22398.011 K x 8.314 J/mol.K=186217.0635 J/mol.
Ea=186.217 kJ/mol ~ 186.22 kJ/mol.
(b) The Arrhenius equation also write as
ln(k2/k1)=(-Ea/R)(1/T2-1/T1)
Given T1=283° C=283+273=556 K and k1=3.52x10-7 M-1s-1.
and T2=508° C=508+273=781 K and k2=3.95x10-2 M-1s-1.
ln[(3.95x10-2 M-1s-1)/(3.52x10-7 M-1s-1)]=(-Ea/8.314J/mol.K)[(1/781K)-(1/556K)]
11.628=(-Ea/8.314J/mol.K)(-0.00051815 K-1)
Ea=186577.1 J/mol=186.58 kJ/mol.
(c) Now T1=283°C=556 K, k1=3.52x10-7 M-1s-1. and T2=293°C=293+273=566 K k2=? and Ea=186577.1 J/mol
lnk2/(3.52x10-7 M-1s-1)=(-186577.1 J/mol/8.314 J/mol.K)[(1/566K)-(1/556K)]
lnk2/(3.52x10-7 M-1s-1)=0.71311
k2=(3.52x10-7 M-1s-1)e0.71311
Rate constant at 293°C, k2=7.182x10-7 M-1s-1.
Please let me know if you have any doubt. Thanks.