The rate of disappearance of HBr in the gas phase reaction
2HBr(g) ? $\mathrm { H } _ { 2 }$ (g) + $\mathrm { Br } _ { 2 }$ (g)
Is 0.190 mol L^-1 $s ^ { - 1 }$ at 150 °C.The rate of reaction is ________ mol L^-1 $s ^ { - 1 }$ .

Question

The rate of disappearance of HBr in the gas phase reaction
2HBr(g) ?

\mathrm { H } _ { 2 }

(g) +

\mathrm { Br } _ { 2 }

(g)
Is 0.190 mol L^-1

s ^ { - 1 }

at 150 °C.The rate of reaction is ________ mol L^-1

s ^ { - 1 }

.

Quizplus · Accepted Answer

The rate of reaction is half the rate of disappearance of HBr because the stoichiometry shows 2 moles of HBr produce 1 mole of H2 and 1 mole of Br2. Therefore, the rate of reaction is 0.190 / 2 = 0.095 mol L⁻¹ s⁻¹.

The Rate of Disappearance of HBr in the Gas Phase H2\mathrm { H } _ { 2 }H2​

The Rate of Disappearance of HBr in the Gas Phase $\mathrm { H } _ { 2 }$