what is the rate of formation of Cl2 when the NOCl is reacting at a rate of -0.24 M/s? 2NOCl (g)--> 2NO (g) + Cl2 (g)
In the given reaction, one mole of NOCl produces one mole of Cl2. Therefore, the rate of formation of Cl2 is equal to the rate of disappearance of NOCl.
The rate of formation of Cl2 is given as -0.24 M/s, which indicates that the concentration of NOCl is decreasing at that rate.
Thus, the rate of formation of Cl2 is also -0.24 M/s.
To find the rate of formation of Cl2, we need to consider the stoichiometry of the reaction.
From the balanced equation:
2 NOCl (g) → 2 NO (g) + Cl2 (g)
We can see that for every 2 moles of NOCl reacted, 1 mole of Cl2 is formed. This means that the rate of formation of Cl2 is half the rate of reaction of NOCl.
Given that the rate of reaction of NOCl is -0.24 M/s, the rate of formation of Cl2 would be:
-0.24 M/s ÷ 2 = -0.12 M/s
Therefore, the rate of formation of Cl2 is -0.12 M/s.