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mart [117]
3 years ago
9

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 v

olume 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.
Chemistry
1 answer:
hoa [83]3 years ago
3 0

Answer:

The total energy change, ΔE, in kilojoules = -61.93 kJ

Explanation:

Relationship between ΔH, ΔE and work done is given by first law of thermodynamics.

ΔE = ΔH - PΔV

Where,

ΔH = Change in enthalpy

ΔE = Change in internal energy

PΔV = Work done

Given that,

ΔH = -75.0 kJ = -75000 J

P = 43.0 atm

ΔV = Final volume - initial volume

    = (2.00 - 5.00) = -3.00 L

PΔV = 43 × (-3.00) = -129 L atm

1 L atm = 101.325 J

-129 L atm = 129 × 101.325 = -13071 J

So ,

ΔE = ΔH - PΔV

     = (-75000 J) - ( -13071 J)

     =  -75000 J + 13071 J

     = -61929 J

Total energy change, ΔE = -61.929 kJ

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