Following data is given for the reaction- CaCO3s- CaOs-CO2g - fH-CaOs-635-1kJ mol-1 - fH-CO2g-393-5kJ mol-1 - fH-CaCO3s-1206-9kJ mol-1Predict the effect of temperature on the equilibrium constant of the above reaction-

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Standard XII

Chemistry

Nucleophilic Substitution Reaction

Following dat...

Question

Following data is given for the reaction: $C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
$Δ_{f} H^{⊝} [C a O (s)] = - 635.1 k J$ $m o l^{- 1}$
$Δ_{f} H^{⊝} [C O_{2} (g)] = - 393.5 k J$ $m o l^{- 1}$
$Δ_{f} H^{⊝} [C a C O_{3} (s)] = - 1206.9 k J$ $m o l^{- 1}$
Predict the effect of temperature on the equilibrium constant of the above reaction.

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Solution

$Δ_{r} H^{⊝} = Δ_{f} H^{⊝} [C a O (s)] + Δ_{f} H^{⊝} [C O_{2} (g)] - Δ_{f} H^{⊝} [C a C O_{3} (s)]$
$∴ Δ_{f} H^{⊝} = 178.3 k J$ $m o l^{- 1}$ .
The reaction is endothermic. Hence, according to Le Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

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Q. Follow data is given for the reaction:

C a C O_{3} ⇌ C a O (s) + C O_{2} (g)

Δ_{f} H^{⊖} [C a O (s)] = - 635.1 k J m o l^{- 1}

Δ_{f} H^{⊖} [C O_{2} (g)] = - 393.5 k J m o l^{- 1}

Δ_{f} H^{⊖} [C a C O_{3} (s)] = - 1206.9 k J m o l^{- 1}

Predict the effect of temperature on the equilibrium constant of the above reaction.

Q. Following data is given for the reaction:

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

Δ_{f} H^{o} [C a O (s)] = - 635.1 {kJ mol}^{- 1}

Δ_{f} H^{o} [C O_{2} (g)] = - 393.5 {kJ mol}^{- 1}

Δ_{f} H^{o} [C a C O_{3} (s)] = - 1206.9 {kJ mol}^{- 1}

Predict the effect of temperature on the equilibrium constant of the above reaction.

Standard molar enthalpy of formation, $Δ_{f} H^{⊖}$ is just a special case of enthalpy of reaction, $Δ_{r} H^{⊖}$ . Is the $Δ_{r} H^{⊖}$ for the following reaction same as $Δ_{f} H^{⊖}$ ? Give reason for your answer.
$C a 0 (s) + C O_{2} (g) \to C a C O_{3} (s); Δ_{f} H^{⊖} = - 178.3 k J m o l^{- 1}$

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:

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:

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Following data is given for the reaction: CaCO3(s)→CaO(s)+CO2(g)ΔfH⊝[CaO(s)]=−635.1kJ mol−1ΔfH⊝[CO2(g)]=−393.5kJ mol−1ΔfH⊝[CaCO3(s)]=−1206.9kJ mol−1Predict the effect of temperature on the equilibrium constant of the above reaction.

ΔrH⊝=ΔfH⊝[CaO(s)]+ΔfH⊝[CO2(g)]−ΔfH⊝[CaCO3(s)]∴ΔfH⊝=178.3kJ mol−1.The reaction is endothermic. Hence, according to Le Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.