Question

Find DH for the reaction below, given the following reactions and DH values. 1/2 N2 1/2O2 ® NO

4 NH3(g) 5 O2 (g) ® 4 NO(g) 6 H20 (1),
M DH = -1170 kJ
4 NH3(g) 3 02 (8) ® 2 N2 (g) 6 H20 (1), DH = -1530 kJ

A)2700kj
B)675 kj
C)360 kj
D)90kj

Answer 1

The enthalpy change for the given reaction is -2700 kJ option A).

Step-by-step explanation:

The enthalpy change (ΔH) for the reaction is calculated by summing the enthalpy changes of the individual reactions. We are given the following reactions and their enthalpy changes:

• 1/2 N2 + 1/2 O2 → NO, ΔH = ?
• 4 NH3 + 5 O2 → 4 NO + 6 H2O, ΔH = -1170 kJ
• 4 NH3 + 3 O2 → 2 N2 + 6 H2O, ΔH = -1530 kJ

To find the enthalpy change for the first reaction, we can use the enthalpy changes of the second and third reactions to cancel out the intermediates. Adding the enthalpy changes together gives: ΔH = -1170 kJ + (-1530 kJ) = -2700 kJ.