contestada

The rate of disappearance of HBr in the gas phase reaction 2HBr(g)→H2(g)+Br2(g) is 0.190 Ms−1 at 150 ∘C. The rate of reaction is ________ Ms−1.
(A) 0.0361
(B) 0.0950
(C) 0.0860
(D) 2.63 0.380

Respuesta :

Answer:

Rate of reaction is [tex]0.0950M.s^{-1}[/tex]

Explanation:

  • Applying law of mass action for this reaction: [tex]rate=-\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [H_{2}]}{\Delta t}=\frac{\Delta [Br_{2}]}{\Delta t}[/tex]
  • [tex]-\frac{\Delta [HBr]}{\Delta t}[/tex] represents rate of disappearance of HBr, [tex]\frac{\Delta [H_{2}]}{\Delta t}[/tex] represents rate of appearance of [tex]H_{2}[/tex] and [tex]\frac{\Delta [Br_{2}]}{\Delta t}[/tex] represents rate of appearance of [tex]Br_{2}[/tex]
  • Here, [tex]-\frac{\Delta [HBr]}{\Delta t}=0.190M.s^{-1}[/tex]
  • So, rate of reaction = [tex]\frac{1}{2}\times (0.190M.s^{-1})=0.0950M.s^{-1}[/tex]