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 kJ \\
2 Ca(s)+O_2(g) \rightarrow 2 CaO(s) & \Delta H_2=-1.269 .8 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.
Answer (2)
To find the enthalpy of the overall reaction, manipulate the intermediate reactions.
Reverse the second reaction and divide by 2, which means the enthalpy is halved and its sign is changed.
Add the modified second reaction to the first reaction to obtain the overall reaction.
The enthalpy of the second intermediate equation is halved and has its sign changed. is halved and has its sign changed.
Explanation
Understanding the Problem We are given two intermediate chemical reactions and the final overall chemical equation. Our goal is to determine how the enthalpy of the second intermediate equation is modified when calculating the enthalpy of the overall chemical equation.
Stating the Given Information The given intermediate reactions are:
C a ( s ) + C O 2 ( g ) + 2 1 O 2 ( g ) → C a C O 3 ( s ) with Δ H 1 = − 812.8 k J
2 C a ( s ) + O 2 ( g ) → 2 C a O ( s ) with Δ H 2 = − 1269.8 k J
The final overall chemical equation is:
C a O ( s ) + C O 2 ( g ) → C a C O 3 ( s )
Reversing the Second Reaction To obtain the final equation from the intermediate equations, we need to manipulate the intermediate equations using Hess's Law. First, we reverse the second reaction and divide it by 2:
2 1 [ 2 C a O ( s ) → 2 C a ( s ) + O 2 ( g )] ⇒ C a O ( s ) → C a ( s ) + 2 1 O 2 ( g ) .
The enthalpy change for this reversed reaction is Δ H = − ( 2 1 × − 1269.8 k J ) = 634.9 k J .
Adding the Reactions Next, we add the first reaction to the modified second reaction:
C a ( s ) + C O 2 ( g ) + 2 1 O 2 ( g ) → C a C O 3 ( s ) with Δ H 1 = − 812.8 k J
C a O ( s ) → C a ( s ) + 2 1 O 2 ( g ) with Δ H = 634.9 k J
Adding these two reactions gives:
C a ( s ) + C O 2 ( g ) + 2 1 O 2 ( g ) + C a O ( s ) → C a C O 3 ( s ) + C a ( s ) + 2 1 O 2 ( g )
Simplifying, we get the overall reaction:
C a O ( s ) + C O 2 ( g ) → C a C O 3 ( s )
Calculating the Enthalpy Change The enthalpy change for the overall reaction is the sum of the enthalpy changes of the modified intermediate reactions:
Δ H o v er a ll = Δ H 1 + ( − 2 1 Δ H 2 ) = − 812.8 + 634.9 = − 177.9 k J
Conclusion From the above calculation, we see that to obtain the overall reaction's enthalpy change, the enthalpy of the second reaction ( Δ H 2 ) was halved and its sign was changed.
Examples
Hess's Law, exemplified in this chemical reaction calculation, is crucial in various real-world applications, such as designing efficient industrial processes. For instance, when synthesizing complex compounds, chemists use Hess's Law to calculate the overall energy requirement by breaking down the synthesis into smaller, manageable steps. This approach helps optimize reaction conditions, minimize energy consumption, and reduce costs, making chemical manufacturing more sustainable and economically viable. Understanding these principles also aids in developing new energy sources and improving existing technologies.
The enthalpy of the second intermediate equation is modified by reversing and halving it to obtain the overall reaction. This results in the enthalpy being halved and its sign changing. Thus, the correct option is A.
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