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?
a. 4.50 x 10-4 mol L-l s-l b. 2.25 X 10-4 mol L-l s-l c. 9.00 X 10-4 mol L-l s-l
d. 2.12 x 10-4 mol L-l s-l e. 2.03 x 10-3 mol L-l s-l
HELP!
rate of rxn = 1/2*d(NOBr)/dt = -d(Br2)/dt = -1/2*d(NO)/dt = 4.5E-4
I get
4.50 NOBr x 1Br/2NOBR = 2.25BR
NOBr cancel out and am left with 2.25BR
Is this correct?
yes that's right
To find the rate of consumption of Br2(g), we need to determine the stoichiometry of the reaction. From the balanced equation, we can see that for every 1 mole of Br2, 2 moles of NOBr are formed.
To find the rate of consumption of Br2, we can use the rate law equation:
rate = k [NO][Br2]
Given that the rate of formation of NOBr is 4.50 x 10^-4 mol L^-1 s^-1, we can substitute this value into the rate law equation and solve for [Br2].
4.50 x 10^-4 mol L^-1 s^-1 = k [NO][Br2]
Since we are only interested in the rate of consumption of Br2, we can rearrange the equation to solve for [Br2]:
[Br2] = (4.50 x 10^-4 mol L^-1 s^-1) / (k [NO])
We know that the stoichiometric coefficient of NO is 1 and that the concentration of NO is not given in the question. Therefore, we cannot directly solve for the rate of consumption of Br2 without more information.
Hence, we cannot determine the rate of consumption of Br2 (moles per liter per second) without the value of the rate constant (k) and the concentration of NO.
Therefore, the correct answer is not provided in the options (a, b, c, d, e).