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.
To calculate the delta H for the given reaction, we can use Hess's Law. Hess's Law states that the enthalpy change for a reaction is the same regardless of the route taken.
Here's how we can solve it step by step:
Step 1: Write down the given reactions and their corresponding delta H values:
1. NH3(g) -> 1/2 N2(g) + 3/2 H2(g) delta H = 46 kJ
2. 2 H2(g) + O2(g) -> 2 H2O(g) delta H = -484 kJ
Step 2: Multiply the first equation by 2, the second equation by 3, and reverse the sign of the second equation to balance the number of moles for the reactants and products in the overall reaction:
2 x (1. NH3(g) -> 1/2 N2(g) + 3/2 H2(g))
-3 x (2. 2 H2(g) + O2(g) -> 2 H2O(g))
The resulting equations are:
1. 2 NH3(g) -> N2(g) + 3 H2(g) delta H = 92 kJ
2. -6 H2(g) - 3 O2(g) -> 6 H2O(g) delta H = 1452 kJ
Step 3: Add the balanced equations together:
2 N2(g) + 6 H2O(g) = 3 O2(g) + 4 NH3(g)
Now combine the delta H values:
delta H_total = delta H1 + delta H2
= 92 kJ + (-1452 kJ)
= -1360 kJ
Therefore, the delta H for the reaction: 2 N2(g) + 6 H2O(g) -> 3 O2(g) + 4 NH3(g) is -1360 kJ.