Which of the following statements is correct for the reaction, at constant temperature and pressure?
$CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g)$
a.)$\Delta H = \Delta U$
b.)$\Delta H < \Delta U$
c.)$\Delta H > \Delta U$
d.)None of these
Answer
662.4k+ views
Hint: We have studied the relation between change in enthalpy and change in internal energy as-$\Delta H = \Delta U + \Delta {n_g}RT$
Where $\Delta H$ is the change in enthalpy, $\Delta U$ is the change in internal energy, $\Delta {n_g}$ is the change in gaseous moles.
Complete step by step solution:
We will first study the reaction and then see which option is our answer.
So, the reaction is:-
$CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g)$
To calculate whether the Change in enthalpy is equal to, less than or greater than the change in internal energy, we will see their relationship which is-
$\Delta H = \Delta U + \Delta {n_g}RT$
Let this be the 1st equation.
It can be defined that the Enthalpy of a system is the sum of internal energy and the work required to achieve its pressure and volume.
Further, the equation PV=nRT
We have change in internal energy is-
$\Delta U = nR\Delta T$
From the question, it is clear that Temperature is constant i.e. $\Delta T$=0
$\begin{gathered}
\Delta U = nR\left( 0 \right) \\
\Delta U = 0 \\
\end{gathered} $
Now, from the reaction we can find out value of $\Delta {n_g}$as-
$\Delta {n_g} = 1 - 1 - \dfrac{1}{2} = \left( { - \dfrac{1}{2}} \right)$
Thus, $\Delta H = 0 + \left( { - \dfrac{1}{2}} \right)RT$
$\Delta H = \left( { - \dfrac{1}{2}} \right)RT$
Thus, $\Delta H < \Delta U$
So, the correct answer is “Option B”.
Additional Information:
Enthalpy is an extensive property. It is directly proportional to the size of the system.
The total enthalpy of a system can not be measured directly; instead, we measure the change in enthalpy. For exothermic reactions at constant pressure, the change in enthalpy of a system is equal to the sum of energy released in the reaction, energy retained in the system and energy lost in expansion.
Note: When the temperature is constant; then change in temperature will be zero. The unit for enthalpy is Joule.
Where $\Delta H$ is the change in enthalpy, $\Delta U$ is the change in internal energy, $\Delta {n_g}$ is the change in gaseous moles.
Complete step by step solution:
We will first study the reaction and then see which option is our answer.
So, the reaction is:-
$CO(g) + \dfrac{1}{2}{O_2}(g) \to C{O_2}(g)$
To calculate whether the Change in enthalpy is equal to, less than or greater than the change in internal energy, we will see their relationship which is-
$\Delta H = \Delta U + \Delta {n_g}RT$
Let this be the 1st equation.
It can be defined that the Enthalpy of a system is the sum of internal energy and the work required to achieve its pressure and volume.
Further, the equation PV=nRT
We have change in internal energy is-
$\Delta U = nR\Delta T$
From the question, it is clear that Temperature is constant i.e. $\Delta T$=0
$\begin{gathered}
\Delta U = nR\left( 0 \right) \\
\Delta U = 0 \\
\end{gathered} $
Now, from the reaction we can find out value of $\Delta {n_g}$as-
$\Delta {n_g} = 1 - 1 - \dfrac{1}{2} = \left( { - \dfrac{1}{2}} \right)$
Thus, $\Delta H = 0 + \left( { - \dfrac{1}{2}} \right)RT$
$\Delta H = \left( { - \dfrac{1}{2}} \right)RT$
Thus, $\Delta H < \Delta U$
So, the correct answer is “Option B”.
Additional Information:
Enthalpy is an extensive property. It is directly proportional to the size of the system.
The total enthalpy of a system can not be measured directly; instead, we measure the change in enthalpy. For exothermic reactions at constant pressure, the change in enthalpy of a system is equal to the sum of energy released in the reaction, energy retained in the system and energy lost in expansion.
Note: When the temperature is constant; then change in temperature will be zero. The unit for enthalpy is Joule.
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