Question

reaction 1: \\( 2 \\, \text{o}_3(g) \
ightarrow 3 \\, \text{o}_2(g) \\)

reaction 2: ?

overall reaction: \\( 2 \\, \text{o}_3(g) + 2 \\, \text{no}(g) \
ightarrow 2 \\, \text{no}_2(g) + 2 \\, \text{o}_2(g) \\)

the enthalpy of the overall reaction represented above can be determined by adding the enthalpies of reactions 1 and 2. which of the following could be reaction 2 ?

a \\( \text{no}(g) + \text{o}(g) \
ightarrow \text{no}_2(g) \\)

b \\( (\text{no})_2(g) \
ightarrow 2 \\, \text{no}(g) \\)

c \\( 2 \\, \text{no}(g) + \text{o}_2(g) \
ightarrow 2 \\, \text{no}_2(g) \\)

d \\( 2 \\, \text{no}_2(g) + \text{o}_3(g) \
ightarrow \text{n}_2\text{o}_5(g) + \text{o}_2(g) \\)

Explanation:

Step1: Recall Hess's Law

Hess's Law states that the enthalpy of an overall reaction is the sum of the enthalpies of the individual reactions that add up to the overall reaction. So, we need to find a reaction (Reaction 2) such that when we add Reaction 1 and Reaction 2, we get the overall reaction.

Reaction 1: \( 2 \text{O}_3(g) ightarrow 3 \text{O}_2(g) \)

Overall reaction: \( 2 \text{O}_3(g) + 2 \text{NO}(g) ightarrow 2 \text{NO}_2(g) + 2 \text{O}_2(g) \)

Let's denote Reaction 2 as \( a \text{NO}(g) + b \text{O}_2(g) + \dots ightarrow c \text{NO}_2(g) + \dots \) (we need to find the correct stoichiometry).

Step2: Subtract Reaction 1 from the Overall Reaction

To find Reaction 2, we can perform the operation: Overall Reaction - Reaction 1.

First, write the overall reaction and Reaction 1:

Overall: \( 2 \text{O}_3(g) + 2 \text{NO}(g) ightarrow 2 \text{NO}_2(g) + 2 \text{O}_2(g) \)

Reaction 1: \( 2 \text{O}_3(g) ightarrow 3 \text{O}_2(g) \)

Subtracting Reaction 1 from the overall reaction (which is equivalent to adding the reverse of Reaction 1, but here we just subtract the species):

Left side: \( (2 \text{O}_3 + 2 \text{NO}) - (2 \text{O}_3) = 2 \text{NO} \)

Right side: \( (2 \text{NO}_2 + 2 \text{O}_2) - (3 \text{O}_2) = 2 \text{NO}_2 - \text{O}_2 \) Wait, no, actually, when subtracting reactions, we can think of it as:

Overall reaction = Reaction 1 + Reaction 2

So, Reaction 2 = Overall reaction - Reaction 1

Let's do the species by species:

For \( \text{O}_3 \): 2 in overall, 2 in Reaction 1, so 2 - 2 = 0 (so \( \text{O}_3 \) cancels out)

For \( \text{NO} \): 2 in overall, 0 in Reaction 1, so 2 in Reaction 2 (reactant)

For \( \text{O}_2 \): 2 in overall, 3 in Reaction 1 (but Reaction 1 has \( \text{O}_2 \) as product, so when we subtract Reaction 1, we reverse it? Wait, no, let's use the method of adding reactions.

Let me write Reaction 2 as \( x \text{NO}(g) + y \text{O}_2(g) ightarrow z \text{NO}_2(g) \) (we need to find x, y, z)

Adding Reaction 1 and Reaction 2:

Reaction 1: \( 2 \text{O}_3 ightarrow 3 \text{O}_2 \)

Reaction 2: \( x \text{NO} + y \text{O}_2 ightarrow z \text{NO}_2 \)

Sum: \( 2 \text{O}_3 + x \text{NO} + y \text{O}_2 ightarrow 3 \text{O}_2 + z \text{NO}_2 \)

This should equal the overall reaction: \( 2 \text{O}_3 + 2 \text{NO} ightarrow 2 \text{NO}_2 + 2 \text{O}_2 \)

Now, equate the coefficients:

For \( \text{NO} \): x = 2

For \( \text{NO}_2 \): z = 2

For \( \text{O}_2 \): y + 3 (from Reaction 1's product) = 2 (from overall's product) +? Wait, no, in the sum, the right side is \( 3 \text{O}_2 + z \text{NO}_2 \), and the overall right side is \( 2 \text{NO}_2 + 2 \text{O}_2 \). Wait, I think I messed up the direction. Let's instead write Reaction 2 as a reaction that, when added to Reaction 1, gives the overall reaction.

So:

Reaction 1: \( 2 \text{O}_3 ightarrow 3 \text{O}_2 \)

Reaction 2:?

Sum: \( 2 \text{O}_3 + 2 \text{NO} ightarrow 2 \text{NO}_2 + 2 \text{O}_2 \)

Let's rearrange:

Reaction 2 = Sum - Reaction 1

So, (Sum) - (Reaction 1) is:

\( (2 \text{O}_3 + 2 \text{NO} ightarrow 2 \text{NO}_2 + 2 \text{O}_2) - (2 \text{O}_3 ightarrow 3 \text{O}_2) \)

Which is equivalent to:

\( 2 \text{O}_3 + 2 \text{NO} - 2 \text{O}_3 ightarrow 2 \text{NO}_2 + 2 \text{O}_2 - 3 \text{O}_2 \)

Simplify:

\( 2 \text{NO} ightarrow 2 \text{NO}_2 - \text{O}_2 \)

Wait, that can't be. Wait, no, when subtracting reactions, we can think of it as adding the reverse of Reaction 1. The reverse of Reaction 1 is \( 3 \text{O}_2 ightarrow 2 \text{O}_3 \)

Now, add reverse of Reaction 1 to the overall reaction:

Reverse Reaction 1: \( 3 \text{O}_2 ightarrow 2 \text{O}_3 \)

Overall reaction: \( 2 \text{O}_3 + 2 \text{NO} ightarrow 2 \text{NO}_2 + 2 \text{O}_2 \)

Adding them: \( 3 \text{O}_2 + 2 \text{O}_3 + 2 \text{NO} ightarrow 2 \text{O}_3 + 2 \text{NO}_2 + 2 \text{O}_2 \)

Cancel \( 2 \text{O}_3 \) from both sides:

\( 3 \text{O}_2 + 2 \text{NO} ightarrow 2 \text{NO}_2 + 2 \text{O}_2 \)

Subtract \( 2 \text{O}_2 \) from both sides:

\( \text{O}_2 + 2 \text{NO} ightarrow 2 \text{NO}_2 \)

So Reaction 2 should be \( 2 \text{NO}(g) + \text{O}_2(g) ightarrow 2 \text{NO}_2(g) \), which is option C.

Let's check by adding Reaction 1 and Reaction 2 (option C):

Reaction 1: \( 2 \text{O}_3 ightarrow 3 \text{O}_2 \)

Reaction 2 (option C): \( 2 \text{NO} + \text{O}_2 ightarrow 2 \text{NO}_2 \)

Sum: \( 2 \text{O}_3 + 2 \text{NO} + \text{O}_2 ightarrow 3 \text{O}_2 + 2 \text{NO}_2 \)

Simplify the \( \text{O}_2 \) terms: 3 \( \text{O}_2 \) on right, 1 \( \text{O}_2 \) on left, so subtract 1 \( \text{O}_2 \) from both sides? Wait, no, let's do the stoichiometry:

Left side: \( 2 \text{O}_3 + 2 \text{NO} + \text{O}_2 \)

Right side: \( 3 \text{O}_2 + 2 \text{NO}_2 \)

Now, subtract \( \text{O}_2 \) from both sides:

Left: \( 2 \text{O}_3 + 2 \text{NO} \)

Right: \( 2 \text{O}_2 + 2 \text{NO}_2 \)

Which is exactly the overall reaction: \( 2 \text{O}_3 + 2 \text{NO} ightarrow 2 \text{NO}_2 + 2 \text{O}_2 \)

Yes, that works.

Answer:

C. \( 2 \text{NO}(g) + \text{O}_2(g)
ightarrow 2 \text{NO}_2(g) \)