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To determine the enthalpy change ([tex]\(\Delta H\)[/tex]) for the overall reaction [tex]\(CaO (s) + CO_2 (g) \rightarrow CaCO_3 (s)\)[/tex], we need to use the given intermediate chemical reactions along with some principles of thermochemistry.
Here are the intermediate chemical reactions:
1. [tex]\( Ca (s) + CO_2(g) + \frac{1}{2} O_2 (g) \rightarrow CaCO_3 (s) \)[/tex] with [tex]\(\Delta H_1 = -812 \text{ kJ}\)[/tex]
2. [tex]\( 2 Ca (s) + O_2 (g) \rightarrow 2 CaO (s) \)[/tex] with [tex]\(\Delta H_2 = -1,269.8 \text{ kJ}\)[/tex]
First, let's write the overall reaction we need the enthalpy for:
[tex]\[ CaO (s) + CO_2 (g) \rightarrow CaCO_3 (s) \][/tex]
We can see that the product of the overall reaction, CaCO_3(s), appears as a product in the first intermediate reaction. The reactant CaO(s) appears in the second intermediate reaction. Now, let's manipulate the intermediate reactions to derive the overall reaction.
To use Hess's Law, we need to rearrange and possibly reverse and scale the intermediate reactions. Start by modifying them as such:
1. [tex]\( Ca (s) + CO_2 (g) + \frac{1}{2} O_2 (g) \rightarrow CaCO_3 (s) \)[/tex]
2. Reverse the second reaction to have [tex]\(CaO\)[/tex] on the reactant side:
[tex]\[ 2 CaO (s) \rightarrow 2 Ca (s) + O_2 (g) \][/tex]
When we reverse a reaction, we change the sign of [tex]\(\Delta H\)[/tex]:
[tex]\[ \Delta H = +1269.8 \text{ kJ} \][/tex]
Now, divide the entire reversed reaction by 2, because we need just 1 mole of [tex]\(CaO\)[/tex]:
[tex]\[ CaO (s) \rightarrow Ca (s) + \frac{1}{2} O_2 (g) \][/tex]
Divide [tex]\(\Delta H\)[/tex] by 2 as well:
[tex]\[ \Delta H = \frac{1269.8 \text{ kJ}}{2} = 634.9 \text{ kJ} \][/tex]
Now our intermediate reactions are:
1. [tex]\( Ca (s) + CO_2 (g) + \frac{1}{2} O_2 (g) \rightarrow CaCO_3 (s) \)[/tex] with [tex]\( \Delta H_1 = -812 \text{ kJ}\)[/tex]
2. [tex]\( CaO (s) \rightarrow Ca (s) + \frac{1}{2} O_2 (g) \)[/tex] with [tex]\( \Delta H = +634.9 \text{ kJ}\)[/tex]
Adding these two reactions will give the desired overall reaction:
[tex]\[ Ca (s) + CO_2 (g) + \frac{1}{2} O_2 (g) + CaO (s) \rightarrow CaCO_3 (s) + Ca (s) + \frac{1}{2} O_2 (g) \][/tex]
Cancel out the common terms on both sides:
[tex]\[ CaO (s) + CO_2 (g) \rightarrow CaCO_3 (s) \][/tex]
Add the enthalpies:
[tex]\[ \Delta H = -812 \text{ kJ} + 634.9 \text{ kJ} = -177.1 \text{ kJ} \][/tex]
Therefore:
- Initially, we took [tex]\(\Delta H_2 = -1269.8 \text{ kJ}\)[/tex],
- We reversed the equation which resulted in [tex]\( \Delta H = +1269.8 \text{ kJ}\)[/tex],
- Then, we halved the second reaction to get [tex]\( \Delta H = +634.9 \text{ kJ}\)[/tex].
So, in the calculation of the overall chemical equation's enthalpy, the enthalpy of the second intermediate reaction is halved and has its sign changed.
Here are the intermediate chemical reactions:
1. [tex]\( Ca (s) + CO_2(g) + \frac{1}{2} O_2 (g) \rightarrow CaCO_3 (s) \)[/tex] with [tex]\(\Delta H_1 = -812 \text{ kJ}\)[/tex]
2. [tex]\( 2 Ca (s) + O_2 (g) \rightarrow 2 CaO (s) \)[/tex] with [tex]\(\Delta H_2 = -1,269.8 \text{ kJ}\)[/tex]
First, let's write the overall reaction we need the enthalpy for:
[tex]\[ CaO (s) + CO_2 (g) \rightarrow CaCO_3 (s) \][/tex]
We can see that the product of the overall reaction, CaCO_3(s), appears as a product in the first intermediate reaction. The reactant CaO(s) appears in the second intermediate reaction. Now, let's manipulate the intermediate reactions to derive the overall reaction.
To use Hess's Law, we need to rearrange and possibly reverse and scale the intermediate reactions. Start by modifying them as such:
1. [tex]\( Ca (s) + CO_2 (g) + \frac{1}{2} O_2 (g) \rightarrow CaCO_3 (s) \)[/tex]
2. Reverse the second reaction to have [tex]\(CaO\)[/tex] on the reactant side:
[tex]\[ 2 CaO (s) \rightarrow 2 Ca (s) + O_2 (g) \][/tex]
When we reverse a reaction, we change the sign of [tex]\(\Delta H\)[/tex]:
[tex]\[ \Delta H = +1269.8 \text{ kJ} \][/tex]
Now, divide the entire reversed reaction by 2, because we need just 1 mole of [tex]\(CaO\)[/tex]:
[tex]\[ CaO (s) \rightarrow Ca (s) + \frac{1}{2} O_2 (g) \][/tex]
Divide [tex]\(\Delta H\)[/tex] by 2 as well:
[tex]\[ \Delta H = \frac{1269.8 \text{ kJ}}{2} = 634.9 \text{ kJ} \][/tex]
Now our intermediate reactions are:
1. [tex]\( Ca (s) + CO_2 (g) + \frac{1}{2} O_2 (g) \rightarrow CaCO_3 (s) \)[/tex] with [tex]\( \Delta H_1 = -812 \text{ kJ}\)[/tex]
2. [tex]\( CaO (s) \rightarrow Ca (s) + \frac{1}{2} O_2 (g) \)[/tex] with [tex]\( \Delta H = +634.9 \text{ kJ}\)[/tex]
Adding these two reactions will give the desired overall reaction:
[tex]\[ Ca (s) + CO_2 (g) + \frac{1}{2} O_2 (g) + CaO (s) \rightarrow CaCO_3 (s) + Ca (s) + \frac{1}{2} O_2 (g) \][/tex]
Cancel out the common terms on both sides:
[tex]\[ CaO (s) + CO_2 (g) \rightarrow CaCO_3 (s) \][/tex]
Add the enthalpies:
[tex]\[ \Delta H = -812 \text{ kJ} + 634.9 \text{ kJ} = -177.1 \text{ kJ} \][/tex]
Therefore:
- Initially, we took [tex]\(\Delta H_2 = -1269.8 \text{ kJ}\)[/tex],
- We reversed the equation which resulted in [tex]\( \Delta H = +1269.8 \text{ kJ}\)[/tex],
- Then, we halved the second reaction to get [tex]\( \Delta H = +634.9 \text{ kJ}\)[/tex].
So, in the calculation of the overall chemical equation's enthalpy, the enthalpy of the second intermediate reaction is halved and has its sign changed.
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