Question

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

Answer 1

Rate of reaction is
`0.0950M.s^(-1)`

Step-by-step explanation:

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