Question

The rate of disappearance of HBr in the gas phase reaction 2HBr(g) → H2 (g) I2 (g) is 0.190 M s-1 at 150°C. The rate of appearance of H2 is ________ M s-1. Group of answer choices

Answer 1

Answer:

The rate of appearance of hydrogen gas is 0.095 M/s.

Explanation:

Rate of the reaction is the change in concentration of of any one of the reactants or products per unit time.

$2HBr(g)\rightarrow H_2 (g) I_2 (g)$

Given:

The rate of disappearance of HBr = $-\frac{d[HBr]}{dt}=0.190 M/s$

Rate of the reaction is given by:

$R=-\frac{1}{2}\frac{d[HBr]}{dt}=\frac{1}{1}\frac{d[H_2]}{dt}$

$R=-\frac{1}{2}\frac{d[HBr]}{dt}=\frac{1}{2}\times 0.190 M/s=0.095 M/s$

Rate of appearance of the hydrogen gas:

$\frac{d[H_2]}{dt}=R =0.095 M/s$

The rate of appearance of hydrogen gas is 0.095 M/s.