Question

In: Chemistry

Rate constants for the gas-phase decomposition of hydrogen iodide, 2 HI(g) → H2 (g) + I2...

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?

Solutions

Expert Solution

(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.


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