Question

Heat of formation of ethane at 300 K and constant pressure is - 85.35kJmol^-1 . Find its value at constant volume.
2C (s) +3H 2 (g) =C 2 H 6 (g)​

Answer 1

Answer:

We have,

ΔH=ΔU−RT

-75830 = ΔU−8.314∗300

heatofformationatconstantvolume=−73340Jor−73.34KJ

Answer 2

Answer:

The heat of formation of ethane at constant volume is equal to -80.36KJ.

Explanation:

Given, the equation representing the formation of ethane :-

2C (s)   +  3H₂(g)  $\longrightarrow$ C₂H₆ (g)

At constant pressure,

Given, the heat of formation of ethane, $q_{p}= -85.35\;KJmol^{-1}$

Consider that $q_v$ is the heat of formation at constant volume.

We know that the heat absorbed or released at constant pressure is equal to enthalpy change while heat absorbed or released at constant volume is equal to internal energy change.

$q_p=\triangle H\; and\; q_v = \triangle V$

$\triangle n_g$ is the change in the number of moles of gaseous substance.

From the chemical equation:-

$\triangle n_g =1-3 = -2$

We know that

$\triangle H = \triangle U + \triangle n_g RT$

$\triangle U =\triangle H - \triangle n_g RT$

Substitute the value of R, T, $\triangle n_g$ and $\triangle H$ in above equation;

$\triangle U = -85.35\times 10^3J-(-2mol)(8.314JK^{-1}mol^{-1})(300K)$

$\triangle U =- 85350J+ 4988.4J$

$\triangle U =-80361.6J$

$\triangle U =-80.36KJ$

Therefore, the heat of formation of ethane at constant volume is equal to  -80.36KJ.

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