Hydrazine, N2H4, is a colorless liquid used in rocket fuel. What is the enthalpy change for the process in which hydranzine is formed from its elements?
N2(g) + H2(g) => N2H4(l)
Use the following reactions and enthalpy changes:
N2H4 + O2 => 2H2O DeltaH = -622.2KJ
H2 + 1/2O2 => H2O DeltaH = -285.5KJ
So how would you do this?
To determine the enthalpy change for the formation of hydrazine (N2H4) from its elements (N2 and H2), we need to use Hess's Law and the given reactions and enthalpy changes.
Hess's Law states that the enthalpy change for a reaction is the same regardless of whether it occurs in one step or a series of steps. Therefore, we can use the given reactions to calculate the enthalpy change for the desired reaction.
First, we need to reverse the second reaction given (H2 + 1/2O2 => H2O) since we need H2 as a reactant rather than a product. The enthalpy change (DeltaH) for the reverse reaction will be +285.5 kJ.
Next, we need to multiply the reversed reaction by a factor to balance the number of moles of H2 in the desired reaction. In this case, the reaction is already balanced, so no scaling factor is needed.
Now, we have:
1/2 H2 + 1/4 O2 => 1/2 H2O (reversed second reaction)
Next, we need to add this reversed reaction to the first reaction (N2H4 + O2 => 2H2O) in a way that cancels out the common compounds, leaving us with the desired reaction (N2(g) + H2(g) => N2H4(l)).
Multiplying the reversed second reaction by 2 and adding it to the first reaction gives us:
N2H4 + O2 + 2(H2 + 1/4 O2) => 2H2O + 2(H2O)
N2H4 + O2 + 2H2 + 1/2O2 => 2H2O + 2H2O
N2H4 + O2 + 2H2 + 1/2O2 => 4H2O
Now, we can sum up the enthalpy changes of the reactions on the left side and subtract the enthalpy changes of the reactions on the right side to find the overall enthalpy change.
DeltaH for overall reaction = [DeltaH for (N2H4 + O2 => 2H2O)] + [DeltaH for reversed (H2 + 1/2O2 => H2O)]
DeltaH = -622.2 kJ + (+285.5 kJ) [Note: The reverse reaction has a positive DeltaH]
DeltaH = -622.2 kJ + 285.5 kJ
DeltaH ≈ -336.7 kJ
Therefore, the enthalpy change for the process in which hydrazine (N2H4) is formed from its elements (N2(g) + H2(g)) is approximately -336.7 kJ.
To calculate the enthalpy change for the formation of hydrazine (N2H4) from its elements (N2 and H2), we will use the concept of Hess's Law.
Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states. Therefore, we can calculate the enthalpy change for the desired reaction by combining other known reactions and their enthalpy changes.
The given reactions are:
1) N2H4 + O2 => 2H2O ΔH = -622.2 kJ
2) H2 + 1/2O2 => H2O ΔH = -285.5 kJ
We need to manipulate these reactions to get the desired reaction (N2(g) + H2(g) => N2H4(l)).
First, we need to balance the reactions to make sure the number of atoms on both sides is the same.
1) N2H4 + O2 => 2H2O ΔH = -622.2 kJ
2) 2H2 + O2 => 2H2O ΔH = (-285.5 kJ) * 2 = -571.0 kJ
Now we can reverse the second reaction (multiply it by -1) to obtain:
3) -2H2O => -2H2 - O2 ΔH = +571.0 kJ
By inspecting the desired reaction and the modified reactions, we observe that if we add reaction 1 and reaction 3, we get the desired reaction. Let's add them:
N2H4 + O2 - 2H2O - 2H2O => N2(g) + 3H2O - 2H2 - O2
Simplifying the equation:
N2H4 + O2 - 4H2O => N2(g) + H2(g)
Now we can calculate the overall enthalpy change of the desired reaction using the enthalpy changes of the modified reactions:
ΔH(desired) = ΔH(reaction 1) + ΔH(reaction 3)
= -622.2 kJ + (+571.0 kJ)
= -51.2 kJ
Therefore, the enthalpy change for the formation of hydrazine from its elements is -51.2 kJ.
Your first reaction is not quite complete:
N2H2 + O2 >> N2 + 2H2O deltaH=-622.2kJ/mol
H2+ 1/2 O2>>H2O deltaH=-285.5KJ
Well, if we double the second reaction, and subtract that from the first reaction
N2H2 -2H2 >> N2 deltaH=-622.2+2(285.5)
or
N2H2 >> N2+2H2 deltaH=above about-50kj/mol
reversing the reaction...
N2+H2>>N2N2 deltaH= - above. So, the heat of formation is a +value, an endothermic process.