Internal Energy Video Lessons

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# Problem: A mole of X reacts at a constant pressure of 43.0 atm via the reactionX(g) + 4 Y(g) → 2 Z(g), ΔH° = −75.0 kJBefore the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.Express your answer with the appropriate units.Energy cannot be created nor destroyed, but it can be transferred between a reaction and its surroundings. The change in energy, ΔE, is positive if the reaction absorbs energy, and it is negative if the reaction releases energy. You may also see this expressed in terms of change in internal energy, ΔU. For the purposes of this question, ΔU and ΔE are equal.One way a reaction can transfer energy to or from the surroundings is by releasing or absorbing heat. A reaction can also transfer energy, in the form of work, through a change in volume. The total change in energy is the sum of the heat and work:ΔE=q+wAt constant pressure, q = ΔH and w = −PΔV and soΔE=ΔH-PΔV

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A mole of X reacts at a constant pressure of 43.0 atm via the reaction

X(g) + 4 Y(g) → 2 Z(g), ΔH° = −75.0 kJ

Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.

Energy cannot be created nor destroyed, but it can be transferred between a reaction and its surroundings. The change in energy, ΔE, is positive if the reaction absorbs energy, and it is negative if the reaction releases energy. You may also see this expressed in terms of change in internal energy, ΔU. For the purposes of this question, ΔU and ΔE are equal.

One way a reaction can transfer energy to or from the surroundings is by releasing or absorbing heat. A reaction can also transfer energy, in the form of work, through a change in volume. The total change in energy is the sum of the heat and work:

$\mathbf{\Delta E}\mathbf{=}\mathbf{q}\mathbf{+}\mathbf{w}$

At constant pressure, q = ΔH and w = −PΔV and so

$\mathbf{\Delta E}\mathbf{=}\mathbf{\Delta H}\mathbf{-}\mathbf{P\Delta V}$