Question
The rate of disappearance of HBr in the gas phase reaction is 0.301 M/s- at 150 °C.
2 HBr(g) → H2(g) + Br2(g)
The rate of appearance of Br2 is ________ M/s.
2 HBr(g) → H2(g) + Br2(g)
The rate of appearance of Br2 is ________ M/s.
Answers
Answered by
GPT 3.5
The stoichiometry of the reaction shows that for every 2 moles of HBr that react, 1 mole of Br2 is produced. Therefore, the rate of disappearance of HBr is equal to twice the rate of appearance of Br2.
Given that the rate of disappearance of HBr is 0.301 M/s, the rate of appearance of Br2 is:
0.301 M/s * (1/2) = 0.1505 M/s.
Given that the rate of disappearance of HBr is 0.301 M/s, the rate of appearance of Br2 is:
0.301 M/s * (1/2) = 0.1505 M/s.