In the reaction, `N_(2)+O_(2)hArr2NO`, the moles//litre of `N_(2),O_(2) "and" NO` respectively `0.25,0.05 "and" 1.0` equilibrium, the initial concentration of `N_(2) "and" O_(2)` will respectively be:

Options

A. `0.75 "mol//litre", 0.55 "mole//litre"`
B. `0.50 "mole//litre".0.75 "mole//litre"`
C. `0.25 "mole//litre",0.50 "mole//litre"`
D. `0.25 "mole//litre",1.0 "mole//litre"`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction is:
\[ N_2 + O_2 \rightleftharpoons 2 NO \]

### Step 2: Define initial concentrations.
Let:
- The initial concentration of \( N_2 \) be \( a \) moles/litre.
- The initial concentration of \( O_2 \) be \( b \) moles/litre.
- The initial concentration of \( NO \) is 0 moles/litre (since no product is formed initially).

### Step 3: Define changes in concentration at equilibrium.
Let \( x \) be the amount of \( N_2 \) and \( O_2 \) that reacts to form \( NO \). At equilibrium, the concentrations will be:
- For \( N_2 \): \( a - x \)
- For \( O_2 \): \( b - x \)
- For \( NO \): \( 2x \)

### Step 4: Use the given equilibrium concentrations.
From the problem, we know:
- The equilibrium concentration of \( N_2 \) is \( 0.25 \) moles/litre.
- The equilibrium concentration of \( O_2 \) is \( 0.05 \) moles/litre.
- The equilibrium concentration of \( NO \) is \( 1.0 \) moles/litre.

### Step 5: Set up equations based on equilibrium concentrations.
From the equilibrium concentration of \( NO \):
\[ 2x = 1.0 \]
Thus,
\[ x = 0.5 \]

Now, substituting \( x \) into the equations for \( N_2 \) and \( O_2 \):
1. For \( N_2 \):
\[ a - x = 0.25 \]
\[ a - 0.5 = 0.25 \]
\[ a = 0.25 + 0.5 = 0.75 \]

2. For \( O_2 \):
\[ b - x = 0.05 \]
\[ b - 0.5 = 0.05 \]
\[ b = 0.05 + 0.5 = 0.55 \]

### Step 6: State the initial concentrations.
Thus, the initial concentrations are:
- Initial concentration of \( N_2 \) = \( 0.75 \) moles/litre
- Initial concentration of \( O_2 \) = \( 0.55 \) moles/litre

### Final Answer:
The initial concentrations of \( N_2 \) and \( O_2 \) are \( 0.75 \) moles/litre and \( 0.55 \) moles/litre, respectively.

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