If at a certain instant during the course of the reaction 2 NO2(g) + F2(g) �¨ 2 NO2F(g), the rate of formation of the product were 0.16 mol/L∙s, how fast would the F2 be disappearing (also in mol/L∙s) at that same instant in time?
A) 0.16 mol/L∙s
B) 0.080 mol/L∙s
C) 0.32 mol/L∙s
D) 0.48 mol/L∙s
Wouldn't this be answered the same way as your previous question?
Yes, I got this one :)
Can you explain this one please?
To determine how fast the F2 is disappearing at a certain instant during the reaction, we need to consider the stoichiometry of the reaction.
The balanced equation for the reaction is:
2 NO2(g) + F2(g) -> 2 NO2F(g)
From the balanced equation, we can see that the ratio between the rate of disappearance of F2 and the rate of formation of NO2F is 1:1. In other words, for every 1 mole of NO2F formed, 1 mole of F2 disappears.
Given that the rate of formation of NO2F is 0.16 mol/L∙s at that instant, it means that at the same instant, the rate of disappearance of F2 is also 0.16 mol/L∙s.
Therefore, the correct answer is:
A) 0.16 mol/L∙s