Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
`2NO_((g))+Br_(2(g))hArr2NOBr_((g))`.
When `0.087 "mole"` of `NO` and `0.0437 "mole"` of `Br_(2)` are mixed in a closed container at constant temperature, `0.0518 "mole"` of `NOBr` is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2NO_{(g)} + Br_2_{(g)} \rightleftharpoons 2NOBr_{(g)} \]

### Step 2: Identify initial amounts
We are given:
- Initial moles of \( NO = 0.087 \) moles
- Initial moles of \( Br_2 = 0.0437 \) moles

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the change in moles of \( Br_2 \) that reacts. According to the stoichiometry of the reaction:
- For every 1 mole of \( Br_2 \) that reacts, 2 moles of \( NO \) will react.
- Therefore, the change in moles of \( NO \) will be \( 2x \).

At equilibrium:
- Moles of \( NO \) = Initial moles of \( NO \) - Change in moles of \( NO \)
- Moles of \( Br_2 \) = Initial moles of \( Br_2 \) - Change in moles of \( Br_2 \)
- Moles of \( NOBr \) = Change in moles of \( NOBr \)

### Step 4: Set up the equilibrium expressions
From the problem, we know that at equilibrium, \( 0.0518 \) moles of \( NOBr \) are formed. Since \( 2x = 0.0518 \):
\[ x = \frac{0.0518}{2} = 0.0259 \]

### Step 5: Calculate equilibrium amounts
Now we can calculate the equilibrium amounts of \( NO \) and \( Br_2 \):
- Equilibrium moles of \( NO \):
\[ \text{Moles of } NO = 0.087 - 2(0.0259) = 0.087 - 0.0518 = 0.0352 \text{ moles} \]
- Equilibrium moles of \( Br_2 \):
\[ \text{Moles of } Br_2 = 0.0437 - 0.0259 = 0.0178 \text{ moles} \]

### Final Results
- Equilibrium amount of \( NO \) = \( 0.0352 \) moles
- Equilibrium amount of \( Br_2 \) = \( 0.0178 \) moles

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