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Consider the intermediate chemical reactions.

[tex]\[
\begin{array}{ll}
Ca ( s ) + CO_2( g ) + \frac{1}{2} O_2( g ) \rightarrow CaCO_3( s ) & \Delta H_1 = -812.8 \, \text{kJ} \\
2 Ca ( s ) + O_2( g ) \rightarrow 2 CaO ( s ) & \Delta H_2 = -1,289.8 \, \text{kJ}
\end{array}
\][/tex]

The final overall chemical equation is:

[tex]\[
CaO ( s ) + CO_2( g ) \rightarrow CaCO_3( s )
\][/tex]

When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation

A. is halved and has its sign changed.
B. is halved.
C. has its sign changed.
D. is unchanged.

Sagot :

GinnyAnswer
To determine the enthalpy change ([tex]\( \Delta H \)[/tex]) for the overall reaction, we need to manipulate the given intermediate reactions to match the final reaction. The overall reaction we are aiming for is:
[tex]\[ \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) \][/tex]

Let's examine the given intermediate reactions:

1. [tex]\[ \text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s) \quad \Delta H_1 = -812.8 \, \text{kJ} \][/tex]
2. [tex]\[ 2 \text{Ca} (s) + \text{O}_2 (g) \rightarrow 2 \text{CaO} (s) \quad \Delta H_2 = -1289.8 \, \text{kJ} \][/tex]

For the first intermediate reaction, we notice that it already contains [tex]\(\text{CaCO}_3 (s)\)[/tex] on the product side, which is part of our desired overall reaction. So we can use it as is.

For the second intermediate reaction, we have 2 moles of Ca and 2 moles of CaO. We want to only have 1 mole of CaO, so we will need to reverse half of this reaction to match our desired overall reaction. Here's how we do it:

First, we reverse the reaction:
[tex]\[ 2 \text{CaO} (s) \rightarrow 2 \text{Ca} (s) + \text{O}_2 (g) \][/tex]
When we reverse a reaction, the enthalpy change sign is reversed:
[tex]\[ \Delta H = +1289.8 \, \text{kJ}\][/tex]

Next, we halve the reaction to get:
[tex]\[ \text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g) \][/tex]
Halving the reaction means halving the enthalpy change:
[tex]\[ \Delta H = \frac{1289.8}{2} = 644.9 \, \text{kJ} \][/tex]

Now, we combine the two reactions:

Intermediate reactions:
[tex]\[ \text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s) \quad \Delta H_1 = -812.8 \, \text{kJ} \][/tex]
[tex]\[ \text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g) \quad \Delta H = 644.9 \, \text{kJ} \][/tex]

When we add these intermediate reactions together:
[tex]\[ (\text{Ca} (s) + \text{CO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{CaCO}_3 (s)) + (\text{CaO} (s) \rightarrow \text{Ca} (s) + \frac{1}{2} \text{O}_2 (g)) \][/tex]

The result is:
[tex]\[ \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s)\][/tex]

The enthalpy change for the overall reaction is:
[tex]\[ \Delta H_{\text{overall}} = -812.8 \, \text{kJ} + 644.9 \, \text{kJ} = -167.9 \, \text{kJ} \][/tex]

Thus, the enthalpy change for the overall reaction [tex]\( \text{CaO} (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) \)[/tex] is [tex]\(-167.9 \, \text{kJ}\)[/tex].
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