Question

(a) What is enthalpy change at constant volume? Explain. 

(b) Calculate the enthalpy of transition for carbon from the following: 

C_diamond + O₂ → CO₂ (g)∆H = −94.3 kcal 

C_amorphous + O₂ → CO₂ (g)∆H = −97.6 kcal 

Also calculate the heat required to change. 

1 g of C_diamond to C_amorphous.

Answer 1

(a) When reaction is carried out in a closed vessel so that volume remains constant, i.e., ∆V = 0,then q_p = q_v = ∆V or ∆H = ∆U

(b) Heat of transition of C_diamond to C_amorphous

C_diamond + O₂ →CO₂

∆H = −94.3 kcal …(1)

C_amorphous + O₂ → CO₂

∆H = −97.6 kcal …(2)

Subtract equation (2) from (1)

C_diamond − C_amorphous = −94.3 − (−97.6)

= 3.3 kcal.

This heat represents the transformation of 1 mole C_diamond to C_amorphous

Therefore, heat of transformation for 1 g of

C_diamond to 1g C_amorphous = 3.3 kcal/12

= 0.275 kcal.