The rate of disappearance of HBr in the gas phase reaction 2HBr (g) → H₂(g) + Br₂ (g)
Is 0.190 M s^-1 at 150 °C. The rate of appearance of Br₂ is ________ M s^-1.

Question

The rate of disappearance of HBr in the gas phase reaction 2HBr (g) → H₂(g) + Br₂ (g) Is 0.190 M s^-1 at 150 °C. The rate of appearance of Br₂ is ________ M s^-1.

Quizplus · Accepted Answer

According to the balanced chemical reaction, the stoichiometry is such that 2 moles of HBr react to form 1 mole of Br2. Therefore, the rate of appearance of Br2 is half of the rate of disappearance of HBr.
Rate of appearance of Br2 = 0.190/2 = 0.095 M/s. Answer: (B)

The Rate of Disappearance of HBr in the Gas Phase