Enthalpy of reaction of nitrogen dioxide (use Hess Law)
N2(g) + 2O2(g) → 2NO2(g)
N2 + O2 → 2NO deltaHr= +180 kJmol-1
2NO + O2 → 2NO2 deltaHr= -112 kJmol-1
Starting Point Ending point
N2 (g) + 2O2 (g) → 2NO2 (g)
+180kJ -112kJ
2NO2 + O2 (g)
Use the information to calculate the deltaH and explain how you got your answer.
N2 + O2 → 2NO deltaHr= +180 kJmol-1
2NO + O2 → 2NO2 deltaHr= -112 kJmol-1
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add equation 1 to equation 2 to obtain
N2 + 2O2 ==> 2NO2 which is the equation you want; therefore, add dH for equation 1 to dH for equation 2 which will be 180 kJ + (-112) = ?
That is 68.
N2(g) + 2O2(g) → 2NO2(g) has 2O2.
N2 + O2 → 2NO deltaHr= +180 kJmol-1 does not have 2O2.
What about that?
In the calculations why should I write 2 ΔH (NO2(g)) instead of ΔH (2NO2(g))?
I did this myself after seeing what the teacher had done in class for another calculation in enthalpy.
68 kJ/mol is the correct answer.
You're absolute correctly; i.e., "N2 + O2 → 2NO deltaHr= +180 kJmol-1 does not have 2O2". It has 1 O2 but when you add the O2 from the other equation you now have 2O2 which is what you want. The reason you write 2NO2 is because that's what you get when you add the two equations together. I think your problem is that you aren't add them together. The 2NO from equation 1 cancels with 2NO from equation 2 when you add them together so you are left with N2 + 2O2 ==> 2NO2.
Also, why write 2 ΔH (NO2) instead of ΔH (2NO2)?
Actually I don't know but my best guess is that is 2*delta H (NO2) to show that kJ/mol for NO2 x 2 moles. However, instead of asking a third party (like me) why not ask you teacher why s/he chose to write it that way. I disagree if the teacher means that.
In my opinion the use of kJ/mol should not be used. Technically I think that stands for N2 + O2 ==> 2NO dH = +180 kJ/mol REACTION WHICH MEANS FOR 1 MOL N2 REACTING WITH 1 MOLE O2 TO FORM 2 MOLS OF NO. Many texts, if not most, tend to omit that word "reaction". You can avoid confusion and the need to explain the 2 dH (NO2) notation by writing it this way. N2 + O2 + 180 kJ ==> 2NO. That way it makes it CLEAR that this is an endothermic reaction as well as that the 180 kJ is needed FOR THE REACTION, which just happens to produce 2 mols of NO. Or the notation of N2 + O2 ==> 2NO dH = +180 kJ/mole reaction also avoids confusion. Hope this helps.
To calculate the enthalpy change (ΔH) of the reaction using Hess's Law, we need to manipulate the given equations by multiplying them with appropriate coefficients, if necessary, so that the reactants and products match the desired reaction equation.
Given equations:
1) N2 + O2 → 2NO ΔH = +180 kJ/mol
2) 2NO + O2 → 2NO2 ΔH = -112 kJ/mol
We need to combine these equations to obtain the desired equation:
N2 + 2O2 → 2NO2
Let's manipulate equation 1) by reversing it and multiplying it by 2:
2NO → N2 + 2O2 ΔH = -2 * (+180 kJ/mol) = -360 kJ/mol
Now, equation 2) doesn't have the desired number of NO2 molecules, but notice that the coefficient of NO in equation 2) is half of the desired equation. So, let's multiply equation 2) by 2 to match it with the desired equation:
4NO + 2O2 → 4NO2 ΔH = 2 * (-112 kJ/mol) = -224 kJ/mol
Now, when we sum up the manipulated equations, we obtain the desired equation:
(N2 + 2O2) + (4NO + 2O2) → (2NO2) + (4NO2)
N2 + 6O2 + 4NO → 6NO2
To determine the overall enthalpy change of the reaction, we simply add up the ΔH values of the manipulated equations:
ΔH = (-360 kJ/mol) + (-224 kJ/mol)
ΔH = -584 kJ/mol
Therefore, the enthalpy change (ΔH) of the reaction N2(g) + 2O2(g) → 2NO2(g) using Hess's Law is -584 kJ/mol.