1. Nitric oxide reacts with bromine gas at elevated temperatures according to the equation,
2 NO(g) + Br2(g) = 2 NOBr(g)
The experimental rate law is rate = k[NO][Br2]. In a certain reaction mixture the rate of formation of NOBr(g) was found to be 4.50 x 10-4 mol L-l s-l. What is the rate of consumption of Br2(g), also in mol L-l s-l?
I think that the answer is 9.00 X e-4 because I multiplied 4.50 by 2.
Is this correct?
Wouldn't you divide it by 2?
4.50 NOBr x 1Br/2NOBR = 2.25BR
The NOBr cancel. Is this not correct
Yes it is.
thank you!
To determine the rate of consumption of Br2(g), you need to understand the stoichiometry of the reaction from its balanced equation:
2 NO(g) + Br2(g) = 2 NOBr(g)
The stoichiometric coefficients show that for every 1 mole of Br2 consumed, 2 moles of NOBr are formed. Therefore, the rate of consumption of Br2 will be half of the rate of formation of NOBr.
Given that the rate of formation of NOBr is 4.50 x 10-4 mol L-1 s-1, the rate of consumption of Br2 will be:
4.50 x 10-4 mol L-1 s-1 / 2 = 2.25 x 10-4 mol L-1 s-1
So, the correct answer for the rate of consumption of Br2 is 2.25 x 10-4 mol L-1 s-1, not 9.00 x 10-4 mol L-1 s-1.