Consider the following intermediate chemical equations:
[tex]
\begin{array}{ll}
C(s)+O_2(g) \rightarrow CO_2(g) & \Delta H_1=-393.5 kJ \\
2 CO(g)+O_2(g) \rightarrow 2 CO_2(g) & \Delta H_2=-566.0 kJ \\
2 H_2 O(g) \rightarrow 2 H_2(g)+O_2(g) & \Delta H_3=483.6 kJ
\end{array}
[/tex]
The overall chemical equation is [tex]$C ( s )+ H _2 O ( g ) \rightarrow CO ( g )+ H _2(g)$[/tex]. To calculate the final enthalpy of the overall chemical equation, which step must occur?
A. Reverse the first equation, and change the sign of the enthalpy. Then, add.
B. Reverse the second equation, and change the sign of the enthalpy. Then, add.
C. Multiply the first equation by three, and triple the enthalpy. Then, add.
D. Divide the third equation by two, and double the enthalpy. Then, add.
Answer (1)
Keep the first equation as is.
Reverse the second equation and change the sign of the enthalpy, then divide by 2.
Divide the third equation by 2.
Add the manipulated equations to obtain the overall equation and calculate the final enthalpy change. The step that must occur is to reverse the second equation, and change the sign of the enthalpy. Then, add.
Explanation
Analyzing the Problem We are given three intermediate chemical equations and their enthalpy changes, and we want to find the correct steps to combine them to obtain the overall chemical equation: C ( s ) + H 2 O ( g ) → CO ( g ) + H 2 ( g ) . We will manipulate the given equations to match the overall equation and then determine the corresponding enthalpy change.
Analyzing the First Equation The first equation is C ( s ) + O 2 ( g ) → C O 2 ( g ) with Δ H 1 = − 393.5 k J . We need C ( s ) on the left side, which is already the case. Thus, we keep this equation as is.
Analyzing the Second Equation The second equation is 2 CO ( g ) + O 2 ( g ) → 2 C O 2 ( g ) with Δ H 2 = − 566.0 k J . We need CO ( g ) on the right side. To achieve this, we reverse the equation and divide it by 2: C O 2 ( g ) → CO ( g ) + 2 1 O 2 ( g ) . The enthalpy change becomes − 2 Δ H 2 = − 2 − 566.0 = 283.0 k J .
Analyzing the Third Equation The third equation is 2 H 2 O ( g ) → 2 H 2 ( g ) + O 2 ( g ) with Δ H 3 = 483.6 k J . We need H 2 O ( g ) on the left side and H 2 ( g ) on the right side. To achieve this, we divide the equation by 2: H 2 O ( g ) → H 2 ( g ) + 2 1 O 2 ( g ) . The enthalpy change becomes 2 Δ H 3 = 2 483.6 = 241.8 k J .
Combining the Equations Now, we add the manipulated equations: C ( s ) + O 2 ( g ) → C O 2 ( g ) Δ H 1 = − 393.5 k J C O 2 ( g ) → CO ( g ) + 2 1 O 2 ( g ) Δ H = 283.0 k J H 2 O ( g ) → H 2 ( g ) + 2 1 O 2 ( g ) Δ H = 241.8 k J Adding these gives: C ( s ) + H 2 O ( g ) + O 2 ( g ) + C O 2 ( g ) → C O 2 ( g ) + CO ( g ) + H 2 ( g ) + O 2 ( g ) Simplifying, we get the overall equation: C ( s ) + H 2 O ( g ) → CO ( g ) + H 2 ( g ) .
Determining the Correct Step The overall enthalpy change is Δ H = Δ H 1 − 2 Δ H 2 + 2 Δ H 3 = − 393.5 + 283.0 + 241.8 = 131.3 k J . The step that must occur is to reverse the second equation and change the sign of the enthalpy, then add.
Examples
This concept is used in industrial chemistry to optimize reaction conditions for processes like producing synthetic gas (a mixture of carbon monoxide and hydrogen) from coal and steam. By understanding and manipulating the enthalpy changes of intermediate reactions, chemists can design more efficient and cost-effective processes.
