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# how do you work out this thermodynamics question

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1. The engines of the lunar module of Apollo 11 used methylhydrazine (CH3NHNH2) and dinitrogen tetraoxide. They
react as follows:
4 CH3NHNH2(l) + 5 N2O4(l) → 4 CO2(g) + 12 H2O(l) + 9 N2(g)
Calculate the enthalpy change for the reaction using the following data:
∆Hf: CH3NHNH2(l) = +53, N2O4 (l) = -20, CO2(g) = -393, H2O(l) = -286 kJmol-1
2. I would uses Hess's law, and thus creating an alternate route. This is turn would give me this:

∆H = 4(-53) + 5(20) + 4(-393) + 12(-286) = -5116 kJmol-1

∆H = -5116 kJmol-1

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3. (Original post by starkster29)
The engines of the lunar module of Apollo 11 used methylhydrazine (CH3NHNH2) and dinitrogen tetraoxide. They
react as follows:
4 CH3NHNH2(l) + 5 N2O4(l) → 4 CO2(g) + 12 H2O(l) + 9 N2(g)
Calculate the enthalpy change for the reaction using the following data:
∆Hf: CH3NHNH2(l) = +53, N2O4 (l) = -20, CO2(g) = -393, H2O(l) = -286 kJmol-1
You could try these videos.
4. (Original post by starkster29)
The engines of the lunar module of Apollo 11 used methylhydrazine (CH3NHNH2) and dinitrogen tetraoxide. They
react as follows:
4 CH3NHNH2(l) + 5 N2O4(l) → 4 CO2(g) + 12 H2O(l) + 9 N2(g)
Calculate the enthalpy change for the reaction using the following data:
∆Hf: CH3NHNH2(l) = +53, N2O4 (l) = -20, CO2(g) = -393, H2O(l) = -286 kJmol-1
You could try these videos.

https://exscience.co/blog/a-levelas-...getics-videos/
5. (Original post by gerardhobley)
You could try these videos.
gr8 vids bro

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