Calculate the ∆H for the following reaction: NO + O 🡪 NO2

Question

You are given these three equations:

O2 🡪 2O   ∆H = +495 kJ

2O3 🡪 3O2  ∆H = -427 kJ

NO + O3 🡪 NO2 + O2   ∆H = -199 kJ 

Answer 1

To solve this, we need to treat the approach the same way that we would treat the same type of problem with two different chemical equations, only this time we have three. We will need to manipulate the following three equations in some way in order to get to the final equation:



Equation 1: O2 → 2 O ΔH = + 495 kJ

Equation 2: 2 O3 → 3 O2 ΔH = - 427 kJ

Equation 3: NO + O3 → NO2 + O2 ΔH = - 199 kJ

Final Equation: NO + O → NO2 ΔH = ???? kJ

In order to get to the final equation, we have to possibly multiply or divide equations one, two, and three by themselves, reverse them, or add them together to get the result. To start, let's keep this as simple as possible. We know that we will have exactly one equivalent of NO2 in the products of the final equation and there is one equation that has NO2, equation 3. Due to the fact that no other chemical equations can increase or decrease (cancel out) the amount of NO2 in the products, we have to have one equivalent of NO2 from equation 3. Due to this already being in the given equation, we will keep it as is:

Equation 3: NO + O3 → NO2 + O2 ΔH = - 199 kJ

The next piece is more difficult. We know that we need to get rid of both the O3 and the O2 while adding one equivalent of O. Let's start with adding the one equivalent of O in the reactants. Equation 1 is the only chemical equation with O in it, but it has two equivalents and the final equation we are trying to get to has one. We here must multiply equation 1 by 1/2 to get the one equivalent, reverse it, then add it to equation 3 to get the O in the reactants:

Equation 1: O2 → 2 O ΔH = + 495 kJ

Multiplied equation 1: 1/2 O2 → O ΔH = (+ 495 kJ)*1/2 = +247.5

Multiplied, flipped equation 1: O → 1/2 O2 ΔH = -247.5

Equation 3: NO + O3 → NO2 + O2 ΔH = -199 kJ

Added modified equation 1 + 3: NO + O3 + O → NO2 + O2 + 1/2 O2 ΔH = -247.5 - 199 kJ = -446.5 kJ

Simplified added equations (equation X): NO + O3 + O → NO2 + 3/2 O2 ΔH = -247.5 - 199 kJ = -446.5 kJ

When adding two equations together, we add the ΔH of each equation as well in order to obtain the ΔH of the added equation. When multiplying an equation by a variable, we must also multiply the ΔH by that variable. Now that we have done this, we need to use equation 2 to get rid of the remaining O3 and O2. We see that we have 3/2 equivalent of O2 and one equivalent of O3, allowing us to then flip equation 2 and divide it by two before adding it to the simplified added equations to cancel out those unwanted oxygen species:

Equation 2: 2 O3 → 3 O2 ΔH = - 427 kJ

Flipped equation 2: 3 O2 → 2 O3 ΔH = +427 kJ

Flipped, multiplied equation 2 (equation Y): 3/2 O2 → O3 ΔH = (+427 kJ)*1/2 = +213.5 kJ

Simplified added equations (equation X): NO + O3 + O → NO2 + 3/2 O2 ΔH = -247.5 - 199 kJ = -446.5 kJ

Equations X + Y: NO + O → NO2 ΔH = +213.5 kJ - 446.5 kJ = -233 kJ

When adding two equations together, we add the ΔH of each equation as well in order to obtain the ΔH of the added equation. When multiplying an equation by a variable, we must also multiply the ΔH by that variable. Looking at the addition of the two equations, we were able to obtain the final equation that we were looking for. Therefore, the ΔH that we are looking for is equal to the ΔH for that equation:

Final answer: ΔH = -233 kJ