The amount of Br2 formed in the 15s of the reaction was 0.0153 moles.
Equation of Reaction
\(2HBr \to H_2 + Br_2\)
From the equation of reaction above,
\(-\frac{1}{2}\frac{\delta [HBr]}{[\delta t]} = \frac{\delta [H_2]}{\delta t} = \frac{\delta [Br_2]}{\delta t}\)
In the first 25s of the reaction, the concentration of HBr dropped from 0.520M to 0.486M.
Average Rate of Reaction
\(rate = -\frac{1}{2}(\frac{[HBr]_f - [HBr}_i]}{\delta t}\\rate = -\frac{1}{2}(\frac{0.486-0.520}{2}) \\ rate = 0.00068M/s\)
The average rate of the reaction is 0.00068M/s
Rate of the reaction is
\(rate = \frac{[\delta Br_2]}{\delta t} \\0.00068 = \frac{[\delta Br_2]}{15}\\ \delta [Br_2] = 0.00068 * 15 = 0.0102M\)
The concentration of Br2 in the reaction is 0.0102M.
Number of Moles
The amount of Br2 formed within the first 15s can be calculated as
\(number of moles = molarity * volume\
= 0.0102 * 1.5 = 0.0153mol\)
From the calculations above, 0.0153 moles of Br2 was formed in the first 15s of the reaction.
NB; This was an incomplete question and additional data were sorted and used to solve this problem
Learn more on rate of reaction here;
https://brainly.com/question/19513092
