Question

For the reaction $ \dfrac{1}{8}{S_8}(s) + \dfrac{3}{2}{O_2}(g) \to S{O_3}(g) $ ​ , the difference of heat change at constant pressure and constant volume at 27 degree Celsius will be:
(A) +150 R
(B) -150 R
(C) +450 R
(D) -450 R

Answer 1

Hint: The first law of thermodynamics is simply the application of the conservation of energy principle which states that energy can neither be created nor destroyed, to heat and thermodynamic processes. A way of expressing the first law of thermodynamics in a chemical system is that any change in the internal energy of the system is given by the sum of the heat that flows in and out of the system and the work done on the system by the surroundings.

Complete answer:
We know that the first law of thermodynamics can be written as
 $ \Delta H - \Delta U = \Delta {n_g}RT $
Here the difference of heat change at constant pressure will be denoted by $ \Delta H $ and the difference of heat change at constant volume is denoted by $ \Delta U $ .
Now we need to find the value of $ \Delta {n_g} $ . This is the difference between the number of moles of gaseous reactants and the number of moles of gaseous products.
 $ \Delta {n_g} $ = 1-1.5 = -0.5
Thus we can substitute the values into the equation and thus get:
 $ \Rightarrow \Delta H - \Delta U = - 0.5 \times R \times 300 $
Thus the difference of heat change at constant pressure and constant volume at 27 degree Celsius will be
 $ \Rightarrow \Delta H - \Delta U = - 150R $
Thus the correct answer will be option (B).

Note:
 $ \Delta H $ is known as the heat at constant pressure because at a constant pressure the change in enthalpy is used by the system as the heat flow of the system. This can be easily understood by the formula of enthalpy.
 $ \Delta U $ is known as the heat at constant volume because at constant volume the heat of the reaction is equal to the change in internal energy of the system.