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

Answer:

Rate of reaction is $0.0950M.s^{-1}$

Explanation:

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