Question

For the reaction (at $1240 \,K$ and $1$ atm.)
$ CaCO_3 (s) \rightarrow CaO(s) + CO_2 (g) $
$\Delta H = 176 \, kJ/ mol; \Delta E$ will be :

• A. 160 kJ
• B. 165.6 kJ
• C. 186.4 kJ
• D. 180 kJ

Answer 1

Answer: (B)

For given reaction, $CaCO _{3}( s ) \rightarrow CaO ( s )+ CO _{2}( g )$
$\Delta n _{ g }=1 $
$\Delta H =176\, kJ / mol $
$\Delta E =\Delta H -\Delta nRT $
$\Delta E =176000-(1)(8)(1240) $
$\Delta E =165660\, J $
$\Delta E =165.6\, kJ$