Question

For the reaction, $FeC{{O}_{3}}(s)\to FeO(s)+C{{O}_{2}}(g)$; $\Delta H$= 82.8 kJ at ${{25}^{\circ }}C$. What is $\Delta E$ or $\Delta U$ at ${{25}^{\circ }}C$?
(a)- 82.8 kJ
(b)- 80.32 kJ
(c)- -2394.77 kJ
(d)- 85.28 kJ

Answer 1

Hint: Convert the given temperature to the Kelvin form. The formula that can be used to solve the above question is:
$\Delta U=\Delta H-\Delta nRT$
Here, $\Delta U$ is the change in the internal energy, $\Delta H$ is the change in the enthalpy, $\Delta n$ is the change in a number of moles, R is the gas constant and T is the temperature.

Complete answer:
The given reaction in the question is:
$FeC{{O}_{3}}(s)\to FeO(s)+C{{O}_{2}}(g)$
To find the change in internal energy, we have to use the formula of the first law of thermodynamics, which is:
$\Delta U=\Delta H-\Delta nRT$
Here, $\Delta U$ is the change in the internal energy, $\Delta H$ is the change in the enthalpy, $\Delta n$ is the change in a number of moles, R is the gas constant and T is the temperature.
Given the temperature of the reaction is ${{25}^{\circ }}C$, it has to be converted into Kelvin by adding 273, we get:
T = 273 + 25 = 298 K
The value of gas constant is:
$R=8.314\text{ x 1}{{\text{0}}^{-3}}kJ/mol\text{ }K$
To find the change in the number of moles in the reaction we have to take the difference of the number of moles on the product and the reactant side.
$\Delta n={{n}_{p}}-{{n}_{s}}=1-0=1$
Given the value of change in the enthalpy of the reaction ($\Delta H$) is 82.8 kJ.
Nom, putting the values, we get:
$\Delta U=82.8-(1\text{ x 8}\text{.314 x 1}{{\text{0}}^{-3}}\text{ x 298)}$
$\Delta U=82.8-2.48=80.32\text{ kJ}$
The change in internal energy of the reaction is 80.32 kJ.

Therefore, the correct answer is an option (b)- 80.32 kJ.

Note:
While calculating the change in the number of moles, we have taken 1 on the product side because there is only 1 mole of $C{{O}_{2}}$ in the gaseous form, the rest all are solids. Therefore, only the moles of gaseous components should be considered.