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)

165.6 kJ

Answer 2

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