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  • Following data is given for the reaction: CaCO_{3} (s) \rightarrow CaO_{3} (s) + CO_{2} (g) \Delta _{f}H^{\ominus} [CaO (s)] = -635.1 kJ mol^{-1} \Delta _{f}H^{\ominus } [CO2 (g)] = -393.5 kJ mol^{-1}

Following data is given for the reaction: CaCO_{3} (s) \rightarrow CaO (s) + CO_{2} (g)
\Delta _{f}H^{\ominus} [CaO (s)] = -635.1 kJ mol^{-1}
\Delta _{f}H^{\ominus } [CO_{2} (g)] = -393.5 kJ mol^{-1}
\Delta_{f}H^{\ominus } [CaCO_{3} (s)] = -1206.9 kJ mol^{-1}
Predict the effect of temperature on the Equilibrium constant of the above reaction.

Answers (1)

We know that,

\Delta _{r}H^{\Theta } = \Delta _{f}H^{\Theta }[CO_{2}(g)]-\Delta_{f}H^{\Theta }[CaCO_{3}(s)]

\Delta _{r}H^{\Theta }= 178.3KJmol^{-1}

.Thus As per the Le Chatelier’s Principle on increasing the temperature, the reaction will be shifting in the forward direction.

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