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We can calculate the moles of Br2 using the performance equation for the reactor. Assuming that the reactor is a CSTR and the reaction is first order, the equation is
t = V/F = (Ca0 - Ca) / Ca
We already know t and V
If we know Ca0, we can solve for Ca,
multiplying Ca and V will result to the moles of Br2.

The amount of Br2 formed in the 15s of the reaction was 0.0153 moles.

Equation of Reaction

[tex]2HBr \to H_2 + Br_2[/tex]

From the equation of reaction above,

[tex]-\frac{1}{2}\frac{\delta [HBr]}{[\delta t]} = \frac{\delta [H_2]}{\delta t} = \frac{\delta [Br_2]}{\delta t}[/tex]

In the first 25s of the reaction, the concentration of HBr dropped from 0.520M to 0.486M.

Average Rate of Reaction

[tex]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[/tex]

The average rate of the reaction is 0.00068M/s

Rate of the reaction is

[tex]rate = \frac{[\delta Br_2]}{\delta t} \\0.00068 = \frac{[\delta Br_2]}{15}\\ \delta [Br_2] = 0.00068 * 15 = 0.0102M[/tex]

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

[tex]number of moles = molarity * volume\\n = 0.0102 * 1.5 = 0.0153mol[/tex]

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

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