Question

Deuterium (D or 2H) is an isotope of hydrogen. The molecule D2 undergoes an exchange reaction with ordinary hydrogen, H2, that leads to isotopic equilibrium. D2(g) + H2(g) 2 DH(g) Kp = 1.8 at 298 K If H0rxn for this reaction is 0.64 kJ/mol, calculate Kp at 415 K.

Answer 1

Answer:

Kp2 = 1.936

Explanation:

We are given that;

Initial temperature(T1) = 298 K

Final temperature(T2) = 415 K

Kp1 = 1.8

ΔHrxn° = 0.64 kJ/mol = 640 J/mol

Now, to solve this we'll make use of Van Hoffs equation.

The Van’t Hoff equation is represented as follows:

In (Kp2/Kp1) = (ΔHrxn°/R)[(1/T1) - (1/T2)]

Where,

K1 and K2 are equilibrium constants and T1 is Initial temperature while T2 is Final temperature and ΔHrxn° is enthalpy reaction. R is gas constant which is 8.314 J/mol.k

Thus, plugging in the relevant values, we have;

In (Kp2/1.8) = (640/8.314)[(1/298) - (1/415)]

In (Kp2/1.8) = 0.07282683811

Kp2/1.8 = e^(0.07282683811)

Kp2 = 1.8 x 1.07554 = 1.936

Kp2 = 1.936

Answer 2

Answer:

$Kp_2} =2.0$

Explanation:

$K_{p1}= 1.8$

$K_{p2}= ???$

$T_1= 298K$

$T_2 = 415 K$

$\delta H$ = 0.64 kJ/mol = 640 J/mol

R = 8.314 J/mol.K

Using Van't Hoff Equation:

$In (\frac{Kp_2}{Kp_1} )=(-\frac{DH}{R} *(\frac{1}{T_2}-\frac{1}{T_1} )$

$In (\frac{Kp_2}{1.8} )=(-\frac{640}{8.314} *(\frac{1}{415}-\frac{1}{298} )$

$In (\frac{Kp_2}{1.8} )=0.072827$

$(\frac{Kp_2}{1.8} )= e^{0.0728727}$

$\frac{Kp_2}{1.8} =1.075593605$

$Kp_2} =1.075593605*1.8$

$Kp_2} =1.936068489$

$Kp_2} =2.0$