Problem 16 - Chemistry An Atoms First Approach

Question

For the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g), \Delta H=+90.7 \mathrm{kJ}\). a. What quantity of heat is required to produce 1 mole of mercury by this reaction? b. What quantity of heat is required to produce 1 mole of oxygen gas by this reaction? c. What quantity of heat would be released in the following reaction as written? $$2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s)$$

Step-by-Step Solution

Answer

a. The heat required to produce 1 mole of mercury is +90.7 kJ. b. The heat required to produce 1 mole of oxygen gas is +181.4 kJ. c. The heat released in the reverse reaction is -181.4 kJ.

Step 1: Recall the balanced equation

The balanced equation for the reaction is: $$\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2}\mathrm{O}_{2}(g)$$

Step 2: Use given enthalpy change

The enthalpy change (∆H) for this reaction is given as +90.7 kJ. This is the heat required to decompose one mole of HgO into liquid mercury and oxygen gas. From the balanced equation, 1 mole of Hg corresponds to 1 mole of HgO. Since the amounts are on a 1:1 ratio, the heat required to produce 1 mole of mercury is the same as the given ∆H: Heat required = +90.7 kJ #b. Heat required to produce 1 mole of oxygen gas#

Step 3: Identify the molar ratios

From the balanced equation, we can determine the ratio between the moles of oxygen gas and the moles of mercury oxide: $$\frac{1}{2}\mathrm{O}_{2}(g) : \mathrm{HgO}(s)$$

Step 4: Calculate heat required

As the heat required for the reaction (∆H) is given for the decomposition of 1 mole of HgO, we can find the heat required to produce 1 mole of O₂ by using the molar ratio: \(\frac{moles\ O_{2}}{moles\ HgO}\) = \(\frac{1}{1/2}\) = 2 We can now multiply the given ∆H by this ratio to determine the heat required to produce 1 mole of oxygen gas: Heat required = 2 × (+90.7 kJ) = +181.4 kJ #c. Heat released in the reverse reaction#

Step 5: Reverse the balanced equation

The reverse reaction is: $$2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s)$$

Step 6: Calculate heat released

For the reverse reaction, the enthalpy change will have the opposite sign, since going from products back to reactants should release the same amount of heat that was required to convert reactants into products. Considering the enthalpy of the reverse reaction, and the stoichiometry is given as: $$\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2}\mathrm{O}_{2}(g),\ \Delta H=-90.7 \mathrm{kJ}$$ Since 2 moles of HgO are formed in the reverse reaction, we will multiply the ∆H by 2: Heat released = 2 × (-90.7 kJ) = -181.4 kJ

Key Concepts

Enthalpy Change

Enthalpy change, represented by the symbol ∆H, is a measure of heat energy absorbed or released during a chemical reaction. This value is an extensive property, meaning it depends on the amount of substance reacting. In the context of our textbook exercise, the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2}\mathrm{O}_{2}(g)\), the given enthalpy change of +90.7 kJ indicates that heat is absorbed in the process of decomposing mercury(II) oxide to produce mercury and oxygen. Since the enthalpy change is positive, this is an endothermic reaction, requiring an input of energy.

Heat of Reaction

The heat of reaction corresponds to the enthalpy change for a reaction but is often referred to specifically in terms of the quantity of heat transferred during the process. When asked about the quantity of heat required to produce 1 mole of mercury or oxygen gas from the decomposition of mercury(II) oxide, the solution specifies that it's equivalent to the enthalpy change provided. For one mole of \(\mathrm{Hg}\), the heat required is exactly +90.7 kJ. However, to produce one mole of \(\mathrm{O}_{2}\), the heat required is double, i.e., +181.4 kJ, due to the molar ratio of oxygen to mercury oxide being 1:2. This helps illustrate how the stoichiometry dictates the amount of heat transfer in a chemical process.

Stoichiometry

Stoichiometry is the part of chemistry that deals with calculating the amounts of reactants and products in a chemical reaction. It is based on the law of conservation of mass and the fact that elements are conserved in chemical reactions. In our exercise, to calculate the heat of reaction, we need to understand the molar ratios given by the balanced chemical equation. The reaction shows that producing two moles of mercury from mercury(II) oxide also involves the generation of one mole of oxygen. Conversely, when the reaction is reversed, the stoichiometry shows that combining two moles of mercury with one mole of oxygen releases heat and produces two moles of \(\mathrm{HgO}(s)\), essentially doubling the enthalpy change to -181.4 kJ. This knowledge of stoichiometry is critical for the quantitative assessment of the thermal dynamics in chemical reactions.