calculate enthalpy of H for the reaction
N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l)
Given the reactions
N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ
H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol
H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187.8 kJ
3 answers
You need to check the post carefully. The equation you want is not balanced. I think you have made two typos. I think the left H2O should be H2O2 and I think the right H2) should be H2O
You are defiantly right.
it is suppose to be 2H2O2(l) and 4H2O(l)
and the first equation is suppose to be -622.2 kJ/mol
actually they are all suppose to be kJ/mol, but that was a typo on the exercise.
For a final answer I got -818.2 kJ/mol
i used the first equation as is. then i used the second equation and multiplied it by two and then for the last equation i reversed it and also multiplied it by 2.
it is suppose to be 2H2O2(l) and 4H2O(l)
and the first equation is suppose to be -622.2 kJ/mol
actually they are all suppose to be kJ/mol, but that was a typo on the exercise.
For a final answer I got -818.2 kJ/mol
i used the first equation as is. then i used the second equation and multiplied it by two and then for the last equation i reversed it and also multiplied it by 2.
N2 + 2F2 ---> 2NF3 calculate the standard enthalpy
0 answers