Answer :
To find the heat of formation of [tex]\( SO_2(g) \)[/tex], let's analyze and combine the given reactions systematically.
The given reactions are:
1. [tex]\( S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) + 2x \)[/tex] kJ
2. [tex]\( SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) + y \)[/tex] kJ
To proceed, we will use the principle of Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.
Let's denote the heat of formation of [tex]\( SO_2(g) \)[/tex] as [tex]\( \Delta H_f^{SO_2} \)[/tex].
First, let's rewrite the given reactions to isolate [tex]\( SO_2(g) \)[/tex]:
- Reaction 1: [tex]\( S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) + 2x \)[/tex] kJ (let's consider this equation as it is)
- Reaction 2: [tex]\( SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) + y \)[/tex] kJ (let's reverse this equation to find the heat change for forming [tex]\( SO_2(g) \)[/tex] from [tex]\( SO_3(g) \)[/tex])
Reversing Reaction 2, we get:
[tex]\[SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) - y \text{ kJ} \][/tex]
Adding the reversed Reaction 2 to Reaction 1, we get:
[tex]\[ S(s) + \frac{3}{2} O_2(g) + SO_3(g) \rightarrow SO_3(g) + 2x \text{ kJ} \][/tex]
[tex]\[SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) - y \text{ kJ} \][/tex]
Combining the two reactions, the [tex]\( SO_3(g) \)[/tex] cancels out:
[tex]\[ S(s) + \frac{3}{2} O_2(g) \rightarrow S(s) + \frac{1}{2} O_2(g) + SO_2(g) + 2x - y \text{ kJ} \][/tex]
This simplifies to:
[tex]\[ S(s) + O_2(g) \rightarrow SO_2(g) + (2x - y) \text{ kJ} \][/tex]
Thus, the heat of formation of [tex]\( SO_2(g) \)[/tex] is:
[tex]\[ \Delta H_f^{SO_2} = -2x + y \][/tex]
Therefore, the correct answer is:
A. [tex]\((y - 2x)\)[/tex]
The given reactions are:
1. [tex]\( S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) + 2x \)[/tex] kJ
2. [tex]\( SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) + y \)[/tex] kJ
To proceed, we will use the principle of Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.
Let's denote the heat of formation of [tex]\( SO_2(g) \)[/tex] as [tex]\( \Delta H_f^{SO_2} \)[/tex].
First, let's rewrite the given reactions to isolate [tex]\( SO_2(g) \)[/tex]:
- Reaction 1: [tex]\( S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) + 2x \)[/tex] kJ (let's consider this equation as it is)
- Reaction 2: [tex]\( SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) + y \)[/tex] kJ (let's reverse this equation to find the heat change for forming [tex]\( SO_2(g) \)[/tex] from [tex]\( SO_3(g) \)[/tex])
Reversing Reaction 2, we get:
[tex]\[SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) - y \text{ kJ} \][/tex]
Adding the reversed Reaction 2 to Reaction 1, we get:
[tex]\[ S(s) + \frac{3}{2} O_2(g) + SO_3(g) \rightarrow SO_3(g) + 2x \text{ kJ} \][/tex]
[tex]\[SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) - y \text{ kJ} \][/tex]
Combining the two reactions, the [tex]\( SO_3(g) \)[/tex] cancels out:
[tex]\[ S(s) + \frac{3}{2} O_2(g) \rightarrow S(s) + \frac{1}{2} O_2(g) + SO_2(g) + 2x - y \text{ kJ} \][/tex]
This simplifies to:
[tex]\[ S(s) + O_2(g) \rightarrow SO_2(g) + (2x - y) \text{ kJ} \][/tex]
Thus, the heat of formation of [tex]\( SO_2(g) \)[/tex] is:
[tex]\[ \Delta H_f^{SO_2} = -2x + y \][/tex]
Therefore, the correct answer is:
A. [tex]\((y - 2x)\)[/tex]
