Question

Caco3 =cao+co2 at 977°c∆H=174 kj /mol then ∆E is

Answer 1

Hii dear!

Here's the answer....

Answer 2

Answer: 163607.5 J/mol.

Explanation:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

Relation of $\Delta H$ with $\Delta E$ is given by the formula:

$\Delta H=\Delta E+{\Delta ng}RT$

Where,

$\Delta H$ = enthalpy change = 174 kJ/mol = 174000 J/mol  (1kJ=1000J)

$\Delta E$ = internal energy change= ?

R = Gas constant = $8.314Jmol^{-1}K^{-1}$

T = temperature = $977^oC=977+273=1250K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-0=1$

Putting values in above equation, we get:

$174000=\Delta E+1\times 8.314\times 1250$

$\Delta E=163607.5J/mol$

Thus the internal energy change is 163607.5 J/mol.