Questions: Calculate ΔHrxn for the following reaction: SrO(s) + CO2(g) → SrCO3(s) Use the following reactions and the given values of ΔH for them: Sr(s) + CO2(g) + 1/2 O2(g) → SrCO3(s), ΔH = -819.5 kJ 2 Sr(s) + O2(g) → 2 SrO(s), ΔH = -1096.0 kJ Express the enthalpy change in kilojoules to four significant figures.

Calculate ΔHrxn for the following reaction:
SrO(s) + CO2(g) → SrCO3(s)

Use the following reactions and the given values of ΔH for them:
Sr(s) + CO2(g) + 1/2 O2(g) → SrCO3(s), ΔH = -819.5 kJ
2 Sr(s) + O2(g) → 2 SrO(s), ΔH = -1096.0 kJ

Express the enthalpy change in kilojoules to four significant figures.

Transcript text: Calculate $\Delta H_{\mathrm{rxn}}$ for the following reaction: \[ \mathrm{SrO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{~s}) \] Use the following reactions and the given values of $\Delta H$ for them: \[ \begin{aligned} \mathrm{Sr}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) & \rightarrow \mathrm{SrCO}_{3}(\mathrm{~s}), \quad \Delta H=-819.5 \mathrm{~kJ} \\ 2 \mathrm{Sr}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) & \rightarrow 2 \mathrm{SrO}(\mathrm{~s}), \quad \Delta H=-1096.0 \mathrm{~kJ} \end{aligned} \] Express the enthalpy change in kilojoules to four significant figures.

Solution

Solution Steps

Step 1: Identify the Target Reaction and Given Reactions

The target reaction is: \[ \mathrm{SrO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{~s}) \]

The given reactions are:

$\mathrm{Sr}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{~s})$, with $\Delta H = -819.5 \, \text{kJ}$
$2 \mathrm{Sr}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SrO}(\mathrm{~s})$, with $\Delta H = -1096.0 \, \text{kJ}$

Step 2: Manipulate the Given Reactions

To find $\Delta H_{\mathrm{rxn}}$ for the target reaction, we need to manipulate the given reactions to match the target reaction.

Reverse the second reaction to express the formation of $\mathrm{SrO}$: \[ 2 \mathrm{SrO}(\mathrm{~s}) \rightarrow 2 \mathrm{Sr}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}), \quad \Delta H = +1096.0 \, \text{kJ} \] Divide the entire reaction by 2: \[ \mathrm{SrO}(\mathrm{~s}) \rightarrow \mathrm{Sr}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}), \quad \Delta H = +548.0 \, \text{kJ} \]

Step 3: Combine the Reactions

Add the modified reactions to obtain the target reaction:

$\mathrm{SrO}(\mathrm{~s}) \rightarrow \mathrm{Sr}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$, $\Delta H = +548.0 \, \text{kJ}$
$\mathrm{Sr}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{~s})$, $\Delta H = -819.5 \, \text{kJ}$

Adding these reactions cancels out $\mathrm{Sr}(\mathrm{~s})$ and $\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$, resulting in: \[ \mathrm{SrO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{~s}) \]

The enthalpy change for the target reaction is: \[ \Delta H_{\mathrm{rxn}} = +548.0 \, \text{kJ} - 819.5 \, \text{kJ} = -271.5 \, \text{kJ} \]