posted by Shellby .
Consider the following data:
* NH3(g) yields 1/2 N2(g) + 3/2 H2(g)
delta H = 46kJ
* 2 H2(g) + O2(g) yields 2 H2o(g)
delta H = -484 kJ
Calculate delta H for the reaction:
2 N2(g) + 6 H2O (g) yields 3 O2(g) + 4 NH3(g)
reverse equation 1 and multiply by 4
reverse equation 2 and multiply by 3.
Add the new 1 and 2. When you do this, delta H is multiplied by the multiplier and the sign is changed when reversed.