Question

Calculate the value of $ \Delta U $ for the following reaction:
 $ {C_2}{H_4}(g) + 3{O_2}(g) \to 2C{O_2}(g) + 2{H_2}O(l) $ , $ \Delta H = - 1410.0kJ $ .

Answer 1

Hint: Assume the temperature to be $ 298K $ . Then, calculate the change in number of moles of gaseous reactants and products. Simply apply the equation for calculating change in enthalpy change and manipulate it to calculate change in internal energy $ \Delta U $ .

Complete answer:
The equation for change in enthalpy is given as:
 $ \Delta H = \Delta U + \Delta {n_g}RT $ … $ (i) $
Where, $ \Delta H \to $ change in enthalpy
 $ \Delta U \to $ change in internal energy
 $ R \to $ gas constant $ = 8.314Jmo{l^{ - 1}}{K^{ - 1}} $
 $ T \to $ temperature $ = 298K $ (assuming room temperature)
 $ \Delta {n_g} \to $ number of moles of gaseous products $ - $ number of moles of gaseous reactants
For the given reaction,
 $ \Delta {n_g} = 2 - (1 + 3) = - 2mol $
Negative sign indicates the decrease in number of moles of gaseous molecules in the products.
To calculate the change in internal energy, $ \Delta U $ , we will manipulate the equation $ (i) $ as follows:
 $ \Delta U = \Delta {n_g}RT - \Delta H $
 $ \Delta U = (8.314 \times 298 \times ( - 2)) - ( - 1410 \times {10^3}) $
On simplifying,
 $ \Delta U = - 1410000 + 4955.144 $
 $ \Delta U = - 1405.045kJ $
For the given reaction, $ \Delta H $ is negative which means that the reaction is exothermic in nature. If $ \Delta H $ is positive then the reaction is endothermic in nature and it will absorb the heat from the surroundings.

Additional Information:
For solid and liquid systems, the difference between $ \Delta H $ and $ \Delta U $ is not significant as solids or liquids do not suffer significant volume change upon heating. However, when the system consists of gas, the difference becomes significant. $ \Delta U $ can be measured using a bomb calorimeter. It is made by immersing a steel vessel (bomb) in a water bath. The whole setup is called a calorimeter.

Note:
While calculating $ \Delta {n_g} $ , only take the no of moles of gaseous reactants and products. Skip the values of any other substance which are not in gaseous phase. Notice that the value of gas constant $ R $ is $ 8.314Jmo{l^{ - 1}}{K^{ - 1}} $ and not $ 0.0821Latm{K^{ - 1}}mo{l^{ - 1}} $ . The latter value is used in the ideal gas equation $ (pV = nRT) $ where usually the values of pressure and volume are expressed in $ atm $ and litres respectively.