Question

The heat liberated on complete combustion of 7.8g Benzene is 327 kJ.This heat has been measured at constant volume and at 27 degrees Celsius.Calculate heat of combustion of benzene at constant pressure at 27 degrees Celsius.

Answer 1

Here is your answer ⤵⤵⤵

We have, C6H6(l) + 15/2O2(g) → 6CO2(g) + 3H2O(l)

Therefore, ∆n = 6 – 15/2 = – 3/2

Also, ∆Uper mol = – (327×78)/7.8

= – 3270 kJ

Therefore, ∆H = ∆U + ∆nRT

= – 3270 + (– 3/2)×300×10–3

Or, ∆H = – 3273.735 (C)

HOPE IT HELPS YOU ☺☺ !!

Answer 2

Hii Mate....

here is your answer ⤵⤵

We have, C6H6(l) + 15/2O2(g) → 6CO2(g) + 3H2O(l)

Therefore, ∆n = 6 – 15/2 = – 3/2

Also, ∆Uper mol = – (327×78)/7.8

= – 3270 kJ

Therefore, ∆H = ∆U + ∆nRT

= – 3270 + (– 3/2)×300×10–3

Or, ∆H = – 3273.735 (C)

HOPE IT HELPS YOU !!