Question

Given the following data:
( ) ( ) ( ) ( ) 2 2 2 CO g H O g CO g H g + à ƒ ¢ † ’ + H ( C)
o o à ƒ ¯  „ 25
= -42.0 kJmol-1
Cp (CO) = 26.8 + 6 x 10-3T JK-1mol-1
Cp (H2O) = 30.1 + 10 x 10-3T JK-1mol-1
Cp (H2) = 28.5 + 2 x 10-3T JK-1mol-1
Cp (CO2) = 26.4 + 42 x 10-3T JK-1mol-1
where T is the thermodynamic temperature, calculate the change in enthalpy for the
reaction at 1000 K.

Answer 1

Solution.

$\Delta H(T) = \Delta H(298 K) + \intop \Delta CpdT$

$\Delta Cp = \sum Cp(f)-\sum Cp(i)$

$\Delta Cp = 0.028 \times T -2$

$\Delta H(1000) = -42000 + \frac{7T^2}{500}|^{1000}_{298} - 2T|^{1000}_{298} = -30647.3 \ J = -30.6 \ \frac{kJ}{mol}$

Answer:

$\Delta H(T) = -30.6 \ \frac{kJ}{mol}$