1 mole of NO 1 mole of `O_(3)` are taken in a 10 L vessel and heated. At equilibrium, `50%` of NO (by mass) reacts with `O_(3)` according to the equation :
`NO_((g))+O_(3(g))hArrNO_(2(g))+O_(2(g))`.
What will be the equilibrium constant for this reaction ?

Options

A. (a) 1
B. (b) 2
C. (c) 3
D. (d) 4

Explanation

To solve the problem step by step, we will follow the given information and apply the principles of chemical equilibrium.

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ \text{NO}_{(g)} + \text{O}_3_{(g)} \rightleftharpoons \text{NO}_2_{(g)} + \text{O}_2_{(g)} \]

### Step 2: Identify initial moles and conditions
We start with:
- 1 mole of NO
- 1 mole of O3
- The volume of the vessel is 10 L.

### Step 3: Determine the amount of NO that reacts
It is given that 50% of NO by mass reacts. Since the molar mass of NO is approximately 30 g/mol, 1 mole of NO weighs 30 g. Therefore, 50% of NO by mass is:
\[ 0.5 \times 30 \, \text{g} = 15 \, \text{g} \]
This corresponds to:
\[ \frac{15 \, \text{g}}{30 \, \text{g/mol}} = 0.5 \, \text{moles} \]
Thus, \( x = 0.5 \) moles of NO reacts.

### Step 4: Calculate the moles at equilibrium
At equilibrium:
- Moles of NO = \( 1 - 0.5 = 0.5 \)
- Moles of O3 = \( 1 - 0.5 = 0.5 \)
- Moles of NO2 = \( 0 + 0.5 = 0.5 \)
- Moles of O2 = \( 0 + 0.5 = 0.5 \)

### Step 5: Calculate the concentrations at equilibrium
Concentration is calculated as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]

Calculating the concentrations:
- \[ [\text{NO}] = \frac{0.5 \, \text{moles}}{10 \, \text{L}} = 0.05 \, \text{M} \]
- \[ [\text{O}_3] = \frac{0.5 \, \text{moles}}{10 \, \text{L}} = 0.05 \, \text{M} \]
- \[ [\text{NO}_2] = \frac{0.5 \, \text{moles}}{10 \, \text{L}} = 0.05 \, \text{M} \]
- \[ [\text{O}_2] = \frac{0.5 \, \text{moles}}{10 \, \text{L}} = 0.05 \, \text{M} \]

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{NO}_2][\text{O}_2]}{[\text{NO}][\text{O}_3]} \]

### Step 7: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we calculated:
\[ K_c = \frac{(0.05)(0.05)}{(0.05)(0.05)} \]

### Step 8: Simplify the expression
\[ K_c = \frac{0.0025}{0.0025} = 1 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 1 \]

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