Question

Consider the following reaction in a vessel with a movable piston. $$\mathrm{R}(g)+\mathrm{T}(g) \longrightarrow \mathrm{X}(g)$$ As the reaction takes place, the piston loses \(1072 \mathrm{~J}\) of heat. The piston moves down and the surroundings do \(549 \mathrm{~J}\) of work on the system. What is \(\Delta E\) ?

Short answer

Answer: The change in internal energy (ΔE) for the system during the chemical reaction is -523 J.

Step-by-step solution

Identify the given data

We are given the following data: Heat lost by the system (q) = -1072 J (since the heat is lost, it is negative) Work done on the system (w) = 549 J (since work is done on the system, it is positive)

Apply the First Law of Thermodynamics

Using the First Law of Thermodynamics, we have: ΔE = q + w

Substitute the given values and calculate ΔE

Let's substitute the given values into the equation: ΔE = (-1072 J) + (549 J) ΔE = -523 J

Write the final answer

The change in internal energy (ΔE) for the system during this reaction is -523 J.