Question

Step(1) Given condition, at constant pressure

(∆H°)p = q = -75.0kJ

Step (2) calculation for work done in the reaction.

w = -P∆V = -P(V2-V​​​​1)

Given Value

P = 43.0atm

V​​​​​​2 = final volume = 2.0L

V​​​​​​1 = initial volume = 5.0L

So,

w = -(43.0atm)(2.0L - 5.0L)

w = -43.0atm(3.0L)

w = 129.0atm-L

(note :- 1atm -L = 101.325J)

so,

w = (129.0×101.325J)

w = 13070.925J

(because 1000j = 1kJ)

w = 13.07kJ

Step (3) Calculation for change in international energy.

∆E = q +w

∆E = -75.0kJ + 13.07kJ

∆E = -61.9KJ

( ∆E = negative value)

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