Question

2. When heated above 500 ºC, potassium nitrate decomposes according to the equation below. 4KNO3 2K2O + 2N2 + 5O2A.If oxygen is being formed at a rate of 0.30 M/s, what are the rates of formation of the other products? B.What is the rate at which the potassium nitrate is decomposed?

Answer 1

Answer:

(a) The rate of formation of K2O is 0.12 M/s.

The rate of formation of N2 is also 0.12 M/s

(b) The rate of decomposition of KNO3 is 0.24 M/s

Explanation:

(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.

Rate of formation of O2 is 0.3 M/s

Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s

Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.

Rate of formation of N2 = (2×0.3/5) = 0.12 M/s

(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.

Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s