Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br₂ (g) ⇌ 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.

The equilibrium amount of NO is 0.0352 mol and the equilibrium amount of Br₂ is 0.0178 mol. — To find the equilibrium amounts, we set up an ICE (Initial, Change, Equilibrium) table in terms of moles. The stoichiometry of the reaction 2NO + Br₂ ⇌ 2NOBr shows that 2 moles of NO and 1 mole of Br₂ react to form 2 moles of NOBr. By determining the change in moles of NOBr formed, we can find the corresponding changes for the reactants and subtract them from their initial amounts.

Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per…

Question 9 of 69intermediate🔧 ApplyNumerical3 marks

Nitric oxide reacts with $Br_2$ and gives nitrosyl bromide as per reaction given below:

2NO (g) + Br_2 (g) \rightleftharpoons 2NOBr (g)

When 0.087 mol of NO and 0.0437 mol of $Br_2$ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and $Br_2$ .

Correct Answer

The equilibrium amount of NO is 0.0352 mol and the equilibrium amount of $Br_2$ is 0.0178 mol.

Exercise: EXERCISES | Q: 6.9 | (Page 42)

For More Understanding

Explanation

To find the equilibrium amounts, we set up an ICE (Initial, Change, Equilibrium) table in terms of moles. The stoichiometry of the reaction $2NO + Br_2 \rightleftharpoons 2NOBr$ shows that 2 moles of NO and 1 mole of $Br_2$ react to form 2 moles of NOBr. By determining the change in moles of NOBr formed, we can find the corresponding changes for the reactants and subtract them from their initial amounts.

Solution Steps

Step 1: Write the balanced chemical equation and set up the initial and equilibrium mole relationships.
$2NO (g) + Br_2 (g) \rightleftharpoons 2NOBr (g)$
Initial moles: $0.087$ mol, $0.0437$ mol, $0$ mol Let $2x$ be the moles of NOBr formed at equilibrium. Change in moles: $-2x$ , $-x$ , $+2x$ Equilibrium moles: $(0.087 - 2x)$ , $(0.0437 - x)$ , $2x$
Step 2: Calculate the value of $x$ using the given equilibrium amount of NOBr. Given, equilibrium moles of NOBr = $0.0518$ mol
$2x = 0.0518$ $x = \frac{0.0518}{2} = 0.0259 \text{ mol}$
Step 3: Calculate the equilibrium amount of NO. Equilibrium moles of NO = $0.087 - 2x$
$= 0.087 - 0.0518 = 0.0352 \text{ mol}$
Step 4: Calculate the equilibrium amount of $Br_2$ . Equilibrium moles of $Br_2$ = $0.0437 - x$
$= 0.0437 - 0.0259 = 0.0178 \text{ mol}$

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