For CaCO3(s)leftharpoons CaO(s)+CO2(g) at 927oC,Δ H =176 kJmol; then textE is:

Question

For CaCO3(s)leftharpoons CaO(s)+CO2(g) at 927oC,Δ H =176 kJmol; then textE is: (A) 180 kJ (B) 186.4 kJ (C) 166.0 kJ (D) 160 kJ

Tardigrade Admin · Accepted Answer

CaCO3(g)leftharpoons CaO(s)+CO2(g) (Δ ng=1-0=1) We know that, Δ H=Δ E+Δ ngRT 176=Δ E +1× 8.31× 10-3× (273 +927) ∴ Δ E=176-8.31× 10-31× 1200 =176-9.97 =166.03 textkJ

Q. For $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ at $927^{o} C, Δ H = 176 k J m o l;$ then $E$ is:

180 kJ

186.4 kJ

166.0 kJ

160 kJ

$C a C O_{3} (g) ⇌ C a O (s) + C O_{2} (g)$ $(Δ n_{g} = 1 - 0 = 1)$ We know that, $Δ H = Δ E + Δ n_{g} RT$
$176 = Δ E + 1 \times 8.31 \times 10^{- 3} \times (273 + 927)$
$∴$ $Δ E = 176 - 8.31 \times 10^{- 31} \times 1200$
$= 176 - 9.97$ $= 166.03 kJ$

Q. For CaCO3​(s)⇌CaO(s)+CO2​(g) at 927oC,ΔH=176kJmol; then E is: