A closed system initially containing 2.00 Times 10^-2 M H_2 and 4.00 Times 10^-2

Question

A closed system initially containing 2.00 Times 10^-2 M H_2 and 4.00 Times 10^-2 M I-2 at 448 degree C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 2.87 Times 10^-2M. Calculate K_c at 448 degree C for the reaction taking place, which is H_2(g) + I_2(g) 2 HI(g) At 298 K, k_1 = 1.36 Times 10^-7 sec^-1. At 323 K, k_2 = 2.72 Times 10^-6 sec^-1. Calculate the energy of activation for this reaction. The rate-determining step is the step in a reaction mechanism.

Explanation / Answer

1.

Kc=[HI]^2/[H2][N2]

Kc= (2.87X10^-2)^2/ (2.0X10^-1* 4.0X10^-2)

Kc= 1.029

2.

WE USE THE EQUATION,

ln(k2/k1) = Ea/R(1/T1 - 1/T2)

T1 = 298 K
k1 = 1.37 x 10^-7 s^-1

T2 = 323 K
k2 = 2.72 x 10^-6 s^-1

R = 8.3145 J/K.mol (Gas constant)

ln(2.72 x 10^-6 s^-1 / 1.37 x 10^-7 s^-1) = -12.7

-12.7 = Ea / 8.3145 (1/298 - 1/323)

1.113 = Ea / 8.3145 (2.59 x 10^-4)
Ea = (-12.7 x 8.3145) / 2.59 x 10^-4 = -40.76 x 10^4 J/mol

3.

b) slowest

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