Video: Calculating the Heat of Reaction for the Reaction of NO and O₂ to Form NO₂ Given the Heats of Formation of All Three and the Balanced Chemical Reaction Equation

What is the heat of reaction for the following reaction, given the heats of formation, Δ𝐻 𝑓, for NO₂ (g), O₂ gas, and NO (g) are 34 kJ/mol, 0 kJ/mol, and 90 kJ/mol, respectively? 2NO (g) + O₂ (g) ⟶ 2NO₂ (g). [A] −124 kJ [B] −112 kJ [C] −56 kJ [D] 56 kJ [E] 124 kJ


Video Transcript

What is the heat of reaction for the following reaction, given the heats of formation, Δ𝐻 𝑓, for NO₂ gas, O₂ gas, and NO gas are 34 kilojoules per mole, zero kilojoules per mole, and 90 kilojoules per mole, respectively? 2NO gas plus O₂ gas react to form 2NO₂ gas. A) Negative 124 kilojoules, B) negative 112 kilojoules, C) negative 56 kilojoules, D) 56 kilojoules, or E) 124 kilojoules.

Heat of reaction is another name for change in enthalpy of the system. Our system is the reactants transforming into the products. Because of the energy required for the reaction to occur or because of the energy the reaction gives out, there may be energy transfer with the surroundings. The reaction is said to be exothermic if the energy out exceeds the energy coming in.

If the energy going out is greater than the energy coming in than the change in enthalpy, or the heat of reaction, will have a negative sign. This indicates that the enthalpy of the system has decreased. If the energy in is greater than the energy out, then we’re dealing with an endothermic reaction. For an endothermic reaction, the enthalpy of the system is increasing, so the change in enthalpy for the reaction will be positive.

The next feature in the question is the term heats of formation. A heat, or enthalpy, of formation is the molar enthalpy change when a substance is formed from its elements in their standard states. For each element, a choice has been made to pick one state as the standard. For oxygen, the standard state is O₂ gas. Let’s take, for example, NO₂ nitrogen dioxide. And we’re going to form it in its gaseous form. Nitrogen dioxide is clearly composed of atoms of nitrogen and atoms of oxygen.

The standard state for nitrogen is N₂ gas. And we’ve already seen that the standard state for oxygen is O₂ gas. If we balanced the nitrogen, we end up with the equation for the formation of NO₂. The enthalpy change for this reaction is the enthalpy of formation, or heat of formation, of nitrogen dioxide. And the value, as per the question, is 34 kilojoules per mole of nitrogen dioxide.

The enthalpy of formation of O₂ gas in the question has been given as zero kilojoules per mole. This makes sense because O₂ gas is the standard state of oxygen. The enthalpy of formation of an element in its standard state is clearly zero. A reaction that produces the reactants can have no change in enthalpy.

The last component of our reaction is NO nitrogen monoxide, otherwise known as nitric oxide. This has an enthalpy of formation of 90 kilojoules per mole. This means that per mole of product, it takes more energy to form nitrogen monoxide than it does to form nitrogen dioxide. We can use a little trick here to work out the reaction enthalpy. The change in enthalpy for a reaction is simply the sum of the formation enthalpies for the products minus the sum of the formation enthalpies for the reactants. You can figure this out from the Hess cycle.

We can build a Hess cycle by first converting our reactants into their constituent elements in their standard states. And then, we can convert from those elements into the products. The second part is simply the enthalpy of formation of the products. And the first part is simply the enthalpy of formation of the reactants, but the other way around. We’re converting the reactants to their constituent elements rather than forming the reactants from their constituent elements. So, we take the negative of the enthalpy of formation of the reactants.

Now, we can start calculating our heat of reaction. The units of the answers are units of energy, meaning that the equation is expressing that we’re adding two moles of NO to one mole of O₂ to produce two moles of NO₂. So, to start off with, we take our two moles of product, two moles of NO₂, and multiply by 34 kilojoules per mole of NO₂. And then, we subtract the term for our reactions, two moles of NO multiplied by 90 kilojoules per mole of NO. We can leave out the term for oxygen because we’d be multiplying by zero anyway.

Our mole units cancel, leaving us with 68 kilojoules minus 180 kilojoules. Leaving us with a heat of reaction of negative 112 kilojoules. So, this reaction is exothermic. And the key feature of getting the correct answer was to recognise we’re generating two moles of product, not one. And we subtract the heats of formation of the reactions from those of the products and not the other way around. Our final answer is negative 112 kilojoules.

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