A certain reaction was performed under a constant pressure of 15.0 atm. During the reaction, the volume contracted from 10.75 L to 3.68 L, and 7.33 kJ of heat was given off. What is $\Delta U$ for this reaction? $(1 L atm =101 J)$
Answer (2)
Calculate the change in volume: Δ V = 3.68 L − 10.75 L = − 7.07 L .
Calculate the work done: w = − ( 15.0 atm ) ( − 7.07 L ) = 106.05 L atm .
Convert the work to kJ: w k J = 106.05 L atm × 101 J/L atm /1000 = 10.71 kJ .
Apply the first law of thermodynamics: Δ U = − 7.33 kJ + 10.71 kJ = 3.38 kJ . The final answer is + 3.38 kJ .
Explanation
Problem Analysis We are given a reaction occurring at constant pressure, with a change in volume and heat released. We need to find the change in internal energy, Δ U , for this reaction. We will use the first law of thermodynamics to solve this problem.
Calculate Volume Change First, calculate the change in volume, Δ V , which is the final volume minus the initial volume: Δ V = V 2 − V 1 = 3.68 L − 10.75 L = − 7.07 L The volume contracted, so Δ V is negative.
Calculate Work Done Next, calculate the work done by the system, w , using the formula w = − P Δ V , where P is the constant pressure: w = − P Δ V = − ( 15.0 atm ) ( − 7.07 L ) = 106.05 L atm The work done is positive because the volume contracted, meaning work was done on the system.
Convert Work to kJ Convert the work from L atm to Joules using the given conversion factor, 1 L atm = 101 J :
w J = w × 101 J/L atm = 106.05 L atm × 101 J/L atm = 10711.05 J Then, convert the work from Joules to kJ by dividing by 1000: w k J = 1000 w J = 1000 J/kJ 10711.05 J = 10.71 kJ
Calculate Change in Internal Energy Apply the first law of thermodynamics, which states that the change in internal energy, Δ U , is equal to the heat added to the system, q , plus the work done on the system, w k J :
Δ U = q + w k J We are given that 7.33 kJ of heat was given off, so q = − 7.33 kJ (negative since heat is released). Substitute the given value of q and the calculated value of w k J to find Δ U :
Δ U = − 7.33 kJ + 10.71 kJ = 3.38 kJ
Final Answer The change in internal energy for this reaction is +3.38 kJ.
Examples
Consider a chemical reaction in a closed container with a movable piston. If the reaction causes the volume to decrease under constant atmospheric pressure, work is done on the system. Simultaneously, if the reaction releases heat, this heat must be accounted for in determining the overall change in the system's internal energy. Understanding these energy transformations is crucial in designing efficient chemical processes and predicting their energy requirements or outputs. For instance, in industrial processes, controlling pressure and volume changes can optimize energy usage and minimize waste.
To calculate the change in internal energy ( Δ U ), we used the first law of thermodynamics, where Δ U = q + w . After determining the work done and converting it to kJ, we found that Δ U = + 3.38 kJ for the reaction.
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