Question

For the reaction at 1240 K and 1 atm. Decomposition of $CaC{{O}_{3}}$ has $\Delta H$ value 176 KJ/mol. the $\Delta U$ equals:
A. 165.6 KJ
B. 160.0 KJ
C. 186.4 KJ
D. 180.0 KJ

Answer 1

Hint: There is a relationship between change in enthalpy and change in internal of a chemical reaction and it is as follows.
\[\Delta U=\Delta H-\Delta nRT\]
Here, $\Delta U$ = Change in internal energy
$\Delta H$ = change in enthalpy
$\Delta n$ = change in number of moles of the reactants
R = gas constant
T = Temperature of the reaction

Complete Solution :
- In the question it is given that the change in enthalpy of a reaction at 1240 K is 176 KJ/mol for the decomposition of the calcium carbonate reaction and said to calculate the change in internal energy if the reaction.
- The decomposition of calcium carbonate can be written as follows.
\[CaC{{O}_{3}}\to CaO+C{{O}_{2}}\]
- The change in number of moles of the above reaction is 2-1 = 1.

- By using the below formula we can calculate the change in internal energy of the decomposition of calcium carbonate.
 \[\Delta U=\Delta H-\Delta nRT\]
Here, $\Delta U$ = Change in internal energy
$\Delta H$ = change in enthalpy = 176 KJ/mol
$\Delta n$ = change in number of moles of the reactants = 1
R = gas constant = 8.314 J/mol
T = Temperature of the reaction = 1240 K

- Substitute all the known values in the above formula to get the change in internal energy of the reaction.
\[\begin{align}
  & \Delta U=\Delta H-\Delta nRT \\
 & \Delta U=176-(1\times 8.314\times 1240) \\
 & \Delta U=165691J \\
 & \Delta U=165.6KJ \\
\end{align}\]
- Therefore the change in internal energy of the decomposition of the calcium carbonate is 165.6 KJ.
So, the correct answer is “Option A”.

Note: We can use $\Delta U$ or $\Delta E$ to represent the change in internal energy of a chemical reaction. We can define the change in internal energy of a reaction is the sum of the work done and heat transferred during the reaction.