For CaCO3s- CaOs-CO2g at 977oC- - H-174 kJ-mol- then - E is-

The correct option is B 163.6 kJ Ca CO _ 3(s) ⟶ Ca O (s) +C O _ 2(g) #160; #160; Δn_g=1 0=1 T= 977 #8451;=977+273=1250K Δ E=Δ ...

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Bond Order and Enthalpy

For CaCO3s→...

Question

For $C a {C O}_{3} (s) \to C a O (s) + {C O}_{2} (g)$ at $977^{o} C$ , $Δ H = 174 k J / m o l$ ; then $Δ E$ is:

160 k J

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163.6 k J

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186.4 k J

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180 k J

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

1240

1

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

C a {C O}_{3 (s)} \to C a O_{(s)} + {C O}_{2 (g)}

Δ H = + 180 K J

93, 39

213 J / m o l / K

Δ S_{t o t a l}

{J K}^{- 1}

300 K

C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)

Δ H

Δ U

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