Question

Ammonia and oxygen react at high temperature as in reaction, 
4HN₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) 
If rate of formation of NO is 3.6 x 10^–3 mol L^–1 .sec^–1 . Calculate the rate of formation of water.

• A. 6.0 x 10^–3 mol L^–1 sec^–1
• B. 3.6 x 10^–3 mol L^–1 sec^–1
• C. 1.8 x 10^–3 mol L^–1 sec^–1
• D. 5.4 x 10^–3 mol L^–1 sec^–1

Answer 1

The Correct Option is D

Solution:

The given reaction is:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

The problem asks to calculate the rate of formation of water, given that the rate of formation of NO is 3.6 x 10⁻³ mol L⁻¹ sec⁻¹.

Step 1: Understanding the Stoichiometric Ratio

From the balanced equation, we can see that for every 4 moles of NO formed, 6 moles of H₂O are formed. This is a stoichiometric ratio of 4:6 or 2:3. This means that for every 4 moles of NO, 6 moles of H₂O will be formed.

Step 2: Relating the Rate of Formation of NO to H₂O

The rate of formation of NO is given as 3.6 x 10⁻³ mol L⁻¹ sec⁻¹. Since the ratio of NO to H₂O is 2:3, we can calculate the rate of formation of H₂O using the stoichiometric relationship.

Step 3: Calculating the Rate of Formation of Water

We can now calculate the rate of formation of H₂O as follows:

Rate of formation of H₂O = (3.6 x 10⁻³ mol L⁻¹ sec⁻¹) × (6 / 4) = 5.4 x 10⁻³ mol L⁻¹ sec⁻¹

Conclusion:

The rate of formation of water is 5.4 x 10⁻³ mol L⁻¹ sec⁻¹, which corresponds to Option (D).