1. Find the enthalpy of the reaction, C(graphite) + O2 (g) → CO2

1.	1. Find the enthalpy of the reaction, C(graphite) + O2 (g) → CO2 (g) Given, (i) C(graphite)+ ½ O2(g) → CO2(g); ∆H =-110.5 kJ mol-1 (ii) CO(g) + ½ O2 (g) → CO2 (g); ∆H =-283.0 kJ mol-1 2. Melting of ice is a spontaneous process. What are the criteria for spontaneity of a process?
Answer» 1. Consider the reaction, C(grahite) + O₂ (g) → CO₂ (g); ∆H = x CO₂can also be prepared through the following two steps: (i) C(graphite)+ ½ O₂ (g) → CO(g); ∆H =110.5 kJ mol^-1 (ii) CO(g) + ½ O₂(g) → CO₂ (g); ∆H =-283.0 kJ mol^-1 Then by Hess’s law, x = (-110.5+-283.0) kJ = -393.5 kJ 2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration. For a spontaneous process ∆S_Total is +ve. For a nonspontaneous process ∆S_Total is -ve.

1. Find the enthalpy of the reaction, C(graphite) + O2 (g) → CO2 (g) Given, (i) C(graphite)+ ½ O2(g) → CO2(g); ∆H =-110.5 kJ mol-1 (ii) CO(g) + ½ O2 (g) → CO2 (g); ∆H =-283.0 kJ mol-1 2. Melting of ice is a spontaneous process. What are the criteria for spontaneity of a process?

Answer»

1. Consider the reaction,

C(grahite) + O₂ (g) → CO₂ (g); ∆H = x

CO₂can also be prepared through the following two steps:

(i) C(graphite)+ ½ O₂ (g) → CO(g); ∆H =110.5 kJ mol^-1

(ii) CO(g) + ½ O₂(g) → CO₂ (g); ∆H =-283.0 kJ mol^-1

Then by Hess’s law,

x = (-110.5+-283.0) kJ = -393.5 kJ

2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration.

For a spontaneous process ∆S_Total is +ve.

For a nonspontaneous process ∆S_Total is -ve.

Discussion

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