Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions: 1. Surroundings get colder and the system decreases in volume. 2. Surroundings get hotter and the system expands in volume. 3. Surroundings get hotter and the system decreases in volume. 4. Surroundings get hotter and the system does not change in volume. 5. Also assume that the magnitude of the volume and temperature changes are similar among the reactions. 6. Rank from most energy released to most energy absorbed. To rank items as equivalent, overlap them. A mole of X reacts at a constant pressure of 43.0 atm via the reaction X(g)+4Y(g)→2Z(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.

The change in the internal energy of a system is positive if the reaction absorbs energy and negative if the reaction releases energy. For a system to cause an increase in volume, it must have very high energy built up to be released.

1. Surroundings get colder and the system decreases in volume. Here, the surrounding absorbs energy resulting in positive ΔE

2. Surroundings get hotter and the system expands in volume. Here energy is released causing the system to be negative

3. Surroundings get hotter and the system decreases in volume. Although there is a decreased volume, the system is negative because it releases energy

4. Surroundings get hotter and the system does not change in volume. System is negative because it releases energy even thgoygh there is no change in volume

Therefore the order from the Most energy released to most Energy Absorbed Is given as 2---> 4--->,3-->---> 1

b) Using

ΔE = q+ w from 1st law of thermodynamics

ΔE= ΔH - P ΔV

gIven

ΔH = -75.0KJ

volume= A change from 5.0L TO 2.0L = Final volume - initial volume = 2-5= -3.00L

P= 43.0atm

ΔE= ΔH - P ΔV

P ΔV = 43 atm x -3 = -129L.atm

We first convert L-atm to Joules.

1 L-atm = 101.325 Joules.

129L.atm = 129 x 101.325 = - 13071 J

to KJ becomes

13071/1000 = - 13.071KJ

Recall ΔE= ΔH - P ΔV and putting values

ΔE = -75.0 - (-13.071 KJ)= -75.0 kJ + 13.071 kJ = -61.9 kJ