For the reaction (at 1240 K and 1 atm.) CaCO3 (s) arrow CaO(s) + CO2 (g) Δ H = 176 kJ/ mol; Δ E will be :

Question

For the reaction (at 1240 K and 1 atm.) CaCO3 (s) arrow CaO(s) + CO2 (g) Δ H = 176 kJ/ mol; Δ E will be : (A) 160 kJ (B) 165.6 kJ (C) 186.4 kJ (D) 180 kJ

Tardigrade Admin · Accepted Answer

For given reaction, CaCO 3( s ) arrow CaO ( s )+ CO 2( g ) Δ n g =1 Δ H =176 kJ / mol Δ E =Δ H -Δ nRT Δ E =176000-(1)(8)(1240) Δ E =165660 J Δ E =165.6 kJ

Q. For the reaction (at $1240 K$ and $1$ atm.)
$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
$Δ H = 176 k J / m o l; Δ E$ will be :

160 kJ

165.6 kJ

186.4 kJ

180 kJ

For given reaction, $C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
$Δ n_{g} = 1$
$Δ H = 176 k J / m o l$
$Δ E = Δ H - Δ n RT$
$Δ E = 176000 - (1) (8) (1240)$
$Δ E = 165660 J$
$Δ E = 165.6 k J$

Q. For the reaction (at 1240K and 1 atm.) CaCO3​(s)→CaO(s)+CO2​(g) ΔH=176kJ/mol;ΔE will be :