Q. 123.7( 3 Votes )
Consider the reaction given below :
CaCO3 (s) → CaO (s) + CO2 (g)
Predict the effect of increase in temperature on the equilibrium constant of this reaction.
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
To understand the type of reaction we need to calculate the enthalpy of the forward reaction.
[Standard enthalpy of formation of CaO + Standard enthalpy of formation CO2] – [Standard enthalpy of formation of CaCO3]
= [-635.1kj mol-1– 393.5kj mole-1]– [-1206.9kj mole-1]
= -1028.6kj mole-1 + 1206.9kj mole-1
= 178.3kj mole-1
This above reaction is an endothermic reaction.
If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat.
Hence, this above reaction would be in the forward direction while increasing the temperature.
Rate this question :
Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl2 (g) ⇌CO (g) + Cl2 (g)
(ii) CH4 (g) + 2S2 (g) ⇌CS2 (g) + 2H2S (g)
(iii) CO2 (g) + C (s) ⇌ 2CO (g)
(iv) 2H2 (g) + CO (g) ⇌ CH3OH (g)
(v) CaCO3 (s) ⇌CaO (s) + CO2 (g)
(vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O(g)NCERT - Chemistry Part-I