The ΔHr for the following reaction is shown.
N2H4(l) + O2(g) → 2 H2O(l) + N2(g)
ΔH = -623 kJ mol-1
Given that the ΔHf of H2O(g) is -286 kJ mol-1, calculate the ΔHf of N2H4(l).
My calculations say that the ΔHf of N2H4(l) = 51kJmol ^ -1
Can anyone tell me if this is correct?
The part that was baffling me was the fact that within the equation, H2O is a liquid state, but the given enthalpy is gaseous. Am I just over thinking the problem or is there a reason for this?
N2H4(l) + O2(g) → 2 H2O(l) + N2(g)
ΔH = -623 kJ mol-1
Given that the ΔHf of H2O(g) is -286 kJ mol-1, calculate the ΔHf of N2H4(l).
My calculations say that the ΔHf of N2H4(l) = 51kJmol ^ -1
Can anyone tell me if this is correct?
The part that was baffling me was the fact that within the equation, H2O is a liquid state, but the given enthalpy is gaseous. Am I just over thinking the problem or is there a reason for this?
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Yes 51 is the right answer