The heat of formation of \( \text{SO}_2(g) \) is given by: $$ S(g) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) + 2x \, \text{kcal} $$ $$ SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) + y \, \text{kcal} $$
Show Hint
- \( \frac{2x}{y} \, \text{kcal} \)
- \( x + y \, \text{kcal} \)
- \( y - 2x \, \text{kcal} \)
- \( 2x + y \, \text{kcal} \)
The Correct Option is C
Approach Solution - 1
The heat of formation of \( \text{SO}_2 \) is the heat change when 1 mole of \( \text{SO}_2(g) \) is formed from its elements in their standard states.
By using the given reactions, the heat of formation is found to be \( y - 2x \).
Final Answer: \( y - 2x \, \text{kcal} \).
Approach Solution -2
Statement (I) states that the first ionization energy of Pb (Lead) is greater than that of Sn (Tin). However, this is incorrect. Although Pb and Sn are in the same group (Group 14), lead (Pb) is lower in the periodic table compared to tin (Sn). As we move down a group, the atomic radius increases, and the outermost electrons are farther from the nucleus, leading to a lower ionization energy. Therefore, the first ionization energy of Pb is less than that of Sn, making Statement (I) false.
Step 2: Analyze Statement (II).
Statement (II) states that the first ionization energy of Ge (Germanium) is greater than that of Si (Silicon). This is true. Ge is in Period 4 and Si is in Period 3, so Si has a smaller atomic radius and the outermost electrons are held more tightly by the nucleus, leading to a higher ionization energy. Therefore, the first ionization energy of Ge is greater than that of Si, making Statement (II) true.
Step 3: Conclusion.
Statement (I) is false, and Statement (II) is true.
Final Answer:
\[ \boxed{\text{Statement I is false but Statement II is true.}} \]
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