Consider the following intermediate chemical equations:
[tex]\begin{array}{ll}
CH_4(g) \rightarrow C(s)+2 H_2(g) & \Delta H_1=74.6 kJ \\
CCl_4(g) \rightarrow C(s)+2 Cl_2(g) & \Delta H_2=95.7 kJ \\
H_2(g)+Cl_2(g) \rightarrow 2 HCl(g) & \Delta H_3=-92.3 kJ
\end{array}[/tex]
What is the enthalpy of the overall chemical reaction [tex]CH _4(g)+4 Cl _2(g) \rightarrow CCl _4(g)+4 HCl ( g )[/tex]?
A. -205.7 kJ
B. -113.4 kJ
C. -14.3 kJ
D. 78.0 kJ
Answer (2)
The enthalpy change for the overall reaction C H 4 ( g ) + 4 C l 2 ( g ) → CC l 4 ( g ) + 4 H Cl ( g ) is calculated to be − 205.7 k J . This is determined by manipulating the enthalpy changes of the given intermediate reactions using Hess's Law. Therefore, the correct choice is A. -205.7 kJ.
;
Reverse the second equation and multiply the third equation by 2.
Add the modified equations to obtain the target reaction.
Sum the enthalpy changes: Δ H o v er a ll = 74.6 − 95.7 + 2 ( − 92.3 ) .
The enthalpy of the overall reaction is − 205.7 k J .
Explanation
Analyzing the Problem and Given Data We are given three intermediate chemical equations with their corresponding enthalpy changes, and we want to find the enthalpy change for the overall reaction:
C H 4 ( g ) → C ( s ) + 2 H 2 ( g ) \t Δ H 1 = 74.6 k J
CC l 4 ( g ) → C ( s ) + 2 C l 2 ( g ) \t Δ H 2 = 95.7 k J
H 2 ( g ) + C l 2 ( g ) → 2 H Cl ( g ) \t Δ H 3 = − 92.3 k J
The target reaction is:
C H 4 ( g ) + 4 C l 2 ( g ) → CC l 4 ( g ) + 4 H Cl ( g )
We will use Hess's Law to calculate the enthalpy change for the overall reaction. Hess's Law states that the enthalpy change for a reaction is independent of the pathway taken, so we can manipulate the given equations to obtain the target reaction and sum their enthalpy changes accordingly.
Manipulating the Equations First, we need to reverse the second equation to get CC l 4 as a product:
C ( s ) + 2 C l 2 ( g ) → CC l 4 ( g ) \t Δ H = − 95.7 k J
Next, we need to multiply the third equation by 2 to get 4 moles of H Cl :
2 H 2 ( g ) + 2 C l 2 ( g ) → 4 H Cl ( g ) \t Δ H = 2 × ( − 92.3 ) = − 184.6 k J
Combining the Equations Now, we add the first equation, the reversed second equation, and the multiplied third equation:
C H 4 ( g ) → C ( s ) + 2 H 2 ( g ) \t Δ H 1 = 74.6 k J C ( s ) + 2 C l 2 ( g ) → CC l 4 ( g ) \t Δ H = − 95.7 k J 2 H 2 ( g ) + 2 C l 2 ( g ) → 4 H Cl ( g ) \t Δ H = − 184.6 k J
Adding these equations gives:
C H 4 ( g ) + C ( s ) + 2 C l 2 ( g ) + 2 H 2 ( g ) + 2 C l 2 ( g ) → C ( s ) + 2 H 2 ( g ) + CC l 4 ( g ) + 4 H Cl ( g )
Simplifying, we get the target reaction:
C H 4 ( g ) + 4 C l 2 ( g ) → CC l 4 ( g ) + 4 H Cl ( g )
Calculating the Overall Enthalpy Change Finally, we sum the enthalpy changes for the modified equations to find the enthalpy change for the overall reaction:
Δ H o v er a ll = Δ H 1 + ( − Δ H 2 ) + 2 × Δ H 3 = 74.6 − 95.7 + 2 × ( − 92.3 ) = 74.6 − 95.7 − 184.6 = − 205.7 k J
Final Answer Therefore, the enthalpy change for the overall chemical reaction is − 205.7 k J .
Examples
Enthalpy calculations are crucial in designing chemical reactors and industrial processes. For example, when producing pharmaceuticals or new materials, knowing the enthalpy change helps engineers determine the energy requirements (heating or cooling) to maintain optimal reaction conditions. This ensures efficient production, prevents hazardous situations, and minimizes energy consumption, leading to cost-effective and sustainable chemical manufacturing.
