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

Answer: (D)

5.4 x 10–3 mol L–1 sec–1

Answer 2

From the balanced equation, we can see that the stoichiometric ratio between NO and H2O is 4:6 or 2:3. This means that for every 4 moles of NO formed, 6 moles of H2O will be formed.
Given that the rate of formation of NO is 3.6x10-3 mol L-1 sec-1
We can calculate the rate of formation of water as follows:
Rate of formation of H2O = (3.6x10-3 mol L-1 sec-1) x $\frac {6}{4}$ = 5.4 x 10-3 mol L-1 sec-1
Therefore, the rate of formation of water is 5.4 x 10-3 mol L-1 sec-1, which corresponds to option (D).