For the reaction CO(g) + \(\frac{1}{2}\) O\(_2\)(g) \(\rightarrow\) CO\(_2\)(g) Which one of the statement is correct at constant T and P?
Show Hint
Just count the moles of gases!
If products have fewer moles of gas (\(\Delta n_g \textless 0\)), then \(\Delta H \textless \Delta E\).
If products have more moles of gas (\(\Delta n_g \textgreater 0\)), then \(\Delta H \textgreater \Delta E\).
\(\Delta H\) is independent of physical state of the reactants
Show Solution
Verified By Collegedunia
The Correct Option isB
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta E\) or \(\Delta U\)) for a given gaseous reaction. Step 2: Key Formula or Approach:
The relationship between enthalpy and internal energy for a reaction involving ideal gases is:
\[ \Delta H = \Delta E + \Delta n_g RT \]
where \(\Delta n_g\) is the change in the number of moles of gaseous products and reactants. Step 3: Detailed Explanation:
Calculate \(\Delta n_g\) for the reaction: CO(g) + \(\frac{1}{2}\) O\(_2\)(g) \(\rightarrow\) CO\(_2\)(g)
Moles of gaseous products (\(n_p\)) = 1
Moles of gaseous reactants (\(n_r\)) = 1 + 0.5 = 1.5
\[ \Delta n_g = n_p - n_r = 1 - 1.5 = -0.5 \]
Since \(\Delta n_g\) is negative:
\[ \Delta H = \Delta E - 0.5 RT \]
Because \(R\) and \(T\) are positive, subtracting a positive value from \(\Delta E\) makes \(\Delta H\) smaller than \(\Delta E\).
Therefore, \(\Delta H \textless \Delta E\). Step 4: Final Answer:
The correct relationship is \(\Delta H \textless \Delta E\).