Find (triangle)H for N2H4 + CH4O > C2HO + N2 + 3H2
2NH3 > N2H4 + H2 22.5 kJ
2NH3 > N2 + 3H2 57.5 kJ
CH20 + H2 > CH4O 81.2 kJ
reverse equation 1 and add to equation 2 and add to the reverse of equation 3.
When you reverse and equation change the sign of dH.
To find the enthalpy change (ΔH) for the given reaction, you can use the concept of Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps in the reaction.
Here's how to calculate the enthalpy change (ΔH) for the given reaction using Hess's Law:
Step 1: Write down the given equations and their corresponding enthalpy changes:
a) 2NH3 ⟶ N2H4 + H2 ΔH1 = 22.5 kJ
b) 2NH3 ⟶ N2 + 3H2 ΔH2 = 57.5 kJ
c) CH4O + H2 ⟶ C2HO + 3H2 ΔH3 = 81.2 kJ
Step 2: Manipulate the given equations to match the desired reaction equation:
We need to cancel out any common species on both sides of the equation. In this case, NH3 and H2 appear in both equations (a) and (b), so we need to manipulate equation (b) by multiplying it by 2 and equation (a) by 3:
a) 6NH3 ⟶ 3N2H4 + 3H2 Multiply equation (a) by 3
b) 4NH3 ⟶ 2N2 + 6H2 Multiply equation (b) by 2
Step 3: Add the manipulated equations together to obtain the desired reaction equation:
6NH3 + 4NH3 ⟶ 3N2H4 + 3H2 + 2N2 + 6H2
Simplifying the equation further:
10NH3 ⟶ 3N2H4 + 5H2 + 2N2
Step 4: Add up the corresponding enthalpy changes:
ΔH(total) = ΔH1 + ΔH2 + ΔH3
ΔH(total) = 22.5 kJ + 57.5 kJ + 81.2 kJ
Therefore, the enthalpy change (ΔH) for the given reaction is:
ΔH = 161.2 kJ