An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Reverse Reaction 2 and calculate the new enthalpy change.
Add Reaction 1 and the reversed Reaction 2, calculating the combined enthalpy change.
Reverse Reaction 3 and calculate the new enthalpy change.
Add the result from the previous step and the reversed Reaction 3 to obtain the target reaction's enthalpy change: − 82 kJ/mol .
Explanation
Problem Analysis We are given three reactions with their corresponding enthalpy changes, and we want to find the enthalpy change for the magnesium combustion reaction using Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the path taken, so we can manipulate the given reactions to obtain the target reaction and sum their enthalpy changes accordingly.
Identifying the Target and Given Reactions The target reaction is: M g ( s ) + 1/2 O 2 ( g ) → M g O ( s ) We need to manipulate the given reactions to match this target reaction. The given reactions are:
Reaction 1: M g ( s ) + 2 H Cl ( a q ) → M g C l 2 ( a q ) + H 2 ( g ) , Δ H 1 = − 450 kJ/mol
Reaction 2: M g O ( s ) + 2 H Cl ( a q ) → M g C l 2 ( a q ) + H 2 O ( l ) , Δ H 2 = − 82 kJ/mol
Reaction 3: H 2 ( g ) + 1/2 O 2 ( g ) → H 2 O ( l ) , Δ H 3 = − 286 kJ/mol
Reversing Reaction 2 First, we reverse Reaction 2: M g C l 2 ( a q ) + H 2 O ( l ) → M g O ( s ) + 2 H Cl ( a q ) , Δ H − 2 = − Δ H 2 = 82 kJ/mol
Adding Reaction 1 and Reversed Reaction 2 Next, we add Reaction 1 and the reversed Reaction 2: M g ( s ) + 2 H Cl ( a q ) + M g C l 2 ( a q ) + H 2 O ( l ) → M g C l 2 ( a q ) + H 2 ( g ) + M g O ( s ) + 2 H Cl ( a q ) Simplifying, we get: M g ( s ) + H 2 O ( l ) → H 2 ( g ) + M g O ( s ) The enthalpy change for this reaction is: Δ H 1 + ( − 2 ) = Δ H 1 + Δ H − 2 = − 450 + 82 = − 368 kJ/mol
Reversing Reaction 3 Now, we reverse Reaction 3: H 2 O ( l ) → H 2 ( g ) + 1/2 O 2 ( g ) , Δ H − 3 = − Δ H 3 = 286 kJ/mol
Adding the Simplified Result and Reversed Reaction 3 Finally, we add the result from step 4 and the reversed Reaction 3: M g ( s ) + H 2 O ( l ) → H 2 ( g ) + M g O ( s ) with Δ H 1 + ( − 2 ) = − 368 kJ/mol and H 2 O ( l ) → H 2 ( g ) + 1/2 O 2 ( g ) with Δ H − 3 = 286 kJ/mol Adding these, we get: M g ( s ) + H 2 O ( l ) → M g O ( s ) + H 2 ( g ) H 2 O ( l ) → H 2 ( g ) + 1/2 O 2 ( g ) M g ( s ) + H 2 O ( l ) + H 2 O ( l ) → H 2 ( g ) + M g O ( s ) + H 2 ( g ) + 1/2 O 2 ( g ) Simplifying, we obtain the target reaction: M g ( s ) + 1/2 O 2 ( g ) → M g O ( s ) The enthalpy change for the target reaction is: Δ H = Δ H 1 + ( − 2 ) + Δ H − 3 = − 368 + 286 = − 82 kJ/mol
Final Answer Therefore, the enthalpy of the magnesium combustion reaction is Δ H = − 82 kJ/mol .
Examples
Understanding enthalpy changes is crucial in many real-world applications. For instance, when designing engines or power plants, engineers need to know how much energy is released or absorbed during chemical reactions to optimize efficiency and prevent overheating. Similarly, in the food industry, knowing the enthalpy of combustion helps determine the caloric content of foods, which is essential for nutritional labeling and dietary planning. By applying Hess's law, we can calculate these enthalpy changes even when direct measurement is impossible, enabling us to make informed decisions in various fields.
