what is the standard enthalpy formation of liquid diethylamine (CH3CH2)2 NH?
N2O5(g) + 8CH4(g) ==> 2(CH3CH2)2NH(l) +5H2O(l), delta H= -1103kJ
Standard delta Hs are listed as kJ/mol so divide -1103 by 2.
that gives me -551.5. that is not 1 of the choices listed.
You might have done well to list the choices. It could be units or something else.
-131kj/mol
-1452kl/mol
-421 kj/mol
+131kj/mol
+421kj/mol
OK. I didn't read the problem correctly. You look up in tables for delta H formation and
delta Hrxn = (delta H products)-(delta H reactants). Solve for delta H of the amine.
To determine the standard enthalpy of formation of liquid diethylamine (C4H11N), we need to use 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 of the reaction.
Given the balanced chemical equation:
N2O5(g) + 8CH4(g) ==> 2(CH3CH2)2NH(l) + 5H2O(l) ΔH = -1103 kJ
We can break down the reaction into simpler steps by manipulating the given equation. Let's consider two reactions:
1. Formation of 2 moles of diethylamine (C4H11N) from its constituent elements:
2C(graphite) + 6H2(g) + 2N2(g) → 2(CH3CH2)2NH(l) ΔH1
2. Combustion of 8 moles of methane (CH4) to form carbon dioxide (CO2) and water (H2O):
8CH4(g) + 16O2(g) → 8CO2(g) + 16H2O(l) ΔH2
Since the reaction enthalpy term of the given equation is the sum of the two above reactions, we can use Hess's Law to solve for the enthalpy of formation of diethylamine.
To find ΔH1, we need to know the standard enthalpies of formation for each individual element.
The standard enthalpy of formation for carbon, hydrogen, and nitrogen in their most stable elemental forms (graphite, H2(g), N2(g)) is considered 0 kJ/mol.
Therefore,
ΔH1 = [2 × ΔHf(CH3CH2)2NH(l)] - [2 × 0 + 6 × 0 + 2 × 0]
Now, to find ΔH2, we can refer to the standard enthalpies of combustion for methane (ΔHc(CH4)) and carbon dioxide (ΔHc(CO2)), which are widely available. The standard enthalpy of formation of water, ΔHf(H2O(l)), is also a commonly known value.
Now we can write the equation for the enthalpy of combustion:
8CH4(g) + 16O2(g) → 8CO2(g) + 16H2O(l) ΔH = ΔH2
By using experimental data, we can calculate ΔHc(CH4) and ΔHc(CO2).
Then we can calculate ΔH2 as:
ΔH2 = [8 × ΔHf(CO2(g))] + [16 × ΔHf(H2O(l))] - [8 × ΔHf(CH4(g))]
Now, substituting the known values into the equation:
ΔH2 = [8 × ΔHf(CO2(g))] + [16 × ΔHf(H2O(l))] - [8 × -890 kJ/mol]
Finally, we substitute the values of ΔH1 and ΔH2 into the given equation to determine the enthalpy of formation of diethylamine:
-1103 kJ = ΔH1 + ΔH2
By rearranging the equation, we can solve for ΔHf(CH3CH2)2NH(l), which is the standard enthalpy of formation of liquid diethylamine.