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# 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 kJ Before the reacti...

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 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.

Use the first law of thermodynamics:

E = q + w, where q is heat and w is work

For a gas reaction where the only work involved is a change in volume,, we can rewrite this as:

E = q - PV

For a chemical reaction where we care about changes, we can write:

ΔE = ΔH - Δ(PV),

You have a pressure of 43.0 atm and a change from 5 to 2 L. So

Δ(PV) = 43 (2-5) L = -129 L-atm

And converting L atm to J:

129 x 101.325 = -13071 J

Finally:

ΔE = -75.0 - (-13071) = -75.0 kJ + 13.071 kJ = -61.9 kJ

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