Question

For the real gases reaction 2co(g) + o2(g) gives 2co2(g); h=-560kj. In 10l rigid vessel at 500k, the initial pressure is 70 bar and after the reaction it becomes 40 bar. The change in internal energy is

Answer 1

The change in internal energy is -529.6 kJ

Given-

• ∆H (Change in enthalpy) = -560 kJ
• Volume of vessel = 10 lit
• Temperature = 500 K
• Initial pressure = 70 bar
• Pressure after the reaction = 40 bar

We know that

∆H = ∆U + ∆(PV)

where ∆H is change in enthalpy and ∆U is change in internal energy.

∆H = ∆U + P₂V₂ - P₁V₁ ------------(i)

We know that these gases are deviate from the ideal gas behavior so we can use-

∆H = ∆U + ∆nRT

∆PV = P₂V₂ - P₁V₁

∆PV = 40 × 10 - 70 × 10

∆PV = - 300 L- atm

∆PV = -30.4 kJ

Therefore from equation (i) we get

-560 = ∆U - 30.4 kJ

∆U = -529.6 kJ

Answer 2

Answer:

answer in attachment

or for easy calculation use 0.1 instead of 101.3 so that there is no requirement to convert into joules in first place.