Question

At 300 K the enthalpy change for the following reaction is -2800 kJ mol$^{-1}$ \[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \] What is the \(\Delta U\) for the same reaction at 300 K (kJ mol$^{-1}$)?

• A. +2802.49
• B. -2800.00
• C. -2814.94
• D. +2802.49

Hint:

When calculating the internal energy change \(\Delta U\), remember the formula: \(\Delta U = \Delta H - \Delta n \cdot R \cdot T\), where \(\Delta n\) is the change in moles of gas.

Answer 1

The Correct Option is B

For reactions involving gases, the relation between the enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) is given by: \[ \Delta H = \Delta U + \Delta n \cdot R \cdot T \] Where: - \(\Delta H\) is the enthalpy change - \(\Delta U\) is the internal energy change - \(\Delta n\) is the change in the number of moles of gases - \(R\) is the gas constant (0.0821 L atm mol$^{-1}$ K$^{-1}$) - \(T\) is the temperature in Kelvin For the given reaction: - \(\Delta n = \text{moles of products} - \text{moles of reactants} = 6 \, \text{mol of CO}_2 + 6 \, \text{mol of H}_2O - 1 \, \text{mol of C}_6H_{12}O_6 - 6 \, \text{mol of O}_2 = 6\) - \(\Delta H = -2800 \, \text{kJ/mol}\) Now we use the relation: \[ \Delta U = \Delta H - \Delta n \cdot R \cdot T \] Substitute the values: \[ \Delta U = -2800 - (6 \cdot 0.0821 \cdot 300) = -2800 - 147.78 = -2800.00 \, \text{kJ/mol} \] Thus, the value of \(\Delta U\) is -2800.00 kJ/mol.

Answer 2

Step 1: Understanding the Given Reaction and Data
The reaction is the combustion of glucose:
\( C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \)
The given enthalpy change, \(\Delta H\), is -2800 kJ mol\(^{-1}\) at 300 K.

Step 2: Relationship Between \(\Delta H\) and \(\Delta U\)
The internal energy change \(\Delta U\) is related to enthalpy change \(\Delta H\) by the equation:
\[ \Delta H = \Delta U + \Delta n_g RT \]
Where,
\(\Delta n_g\) = change in number of moles of gases,
\(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\),
\(T = 300\,K\).

Step 3: Calculate \(\Delta n_g\)
Moles of gaseous reactants = 6 (O\(_2\))
Moles of gaseous products = 6 (CO\(_2\))
So,
\(\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 6 - 6 = 0\)

Step 4: Calculate \(\Delta U\)
Since \(\Delta n_g = 0\),
\[ \Delta H = \Delta U + 0 \times RT \Rightarrow \Delta U = \Delta H = -2800 \, \text{kJ mol}^{-1} \]

Step 5: Conclusion
The internal energy change \(\Delta U\) for the reaction at 300 K is -2800.00 kJ mol\(^{-1}\).