Question

The enthalpy of formation of ethane at
298 K and at constant pressure is - 84.68 kJ.
What is the heat of for
mation at constant
volume at this temperature?​

Answer 1

Given:

temperature T = 298 K

enthalpy ΔH = 84.68 kJ

To find:

heat ΔU =?

Explanation:

The chemical reaction of the formation of ethane is

$2C(s) + 3H_2(g) \rightarrow C_2H_6(g)$

So,

Δn =  1 - 3 = - 2 mol

The enthalpy of formation at the constant temperature & constant pressure is

ΔH = ΔU + Δn RT

where

ΔH = change in enthalpy

ΔU = change in internal energy

Δn = change in gas mol

R = constant = 8.314 x $10^{-3$ KJ$K^{-1} mol^{-1$

T = temperature

Put given values in above eq

84.68 = ΔU + (-2) (8.314 x $10^{-3$) (298)

ΔU = 5.0398 KJ

∴ The heat of formation at constant volume is 5.0398 KJ