1. Tardigrade
  2. Question
  3. Chemistry
  4. Δ H for CaCO3(s) → CaO(s) + CO2(g) is 176 kJ mol-1 at 1240 K. The Δ E for the change is equal to

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Q. $\Delta H$ for $CaCO_3\left(s\right) \to CaO\left(s\right) + CO_2\left(g\right)$ is $176\, kJ\, mol^{-1}$ at $1240\, K$. The $\Delta E$ for the change is equal to

Thermodynamics

Solution:

$\Delta H = \Delta E + \Delta nRT$
$\therefore \Delta E= 176 - 1 \times 8.314 \times 8.314 \times 1240 \times 10^{-3}$
$= 165.6 kJ$