The rate of disappearance of HBr in the gas phase

The rate of disappearance of HBr in the gas phase reaction . The rate of reaction is ? Ms-1.
A. 1.39
B. 0.0860
C. 0.180
D. 0.130
E. 0.720
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hysgubwyri3
Step 1
Since the relationship between the rate of disappearance of reactant and rate of reaction is given by rate of disappearance of reactant / reaction coefficient of reactant = rate of reaction
Step 2
Given : rate of disappearance of reactant $HBr=0.360\frac{M}{s}$
And reaction coefficient of reactant $HBr=2$
$⇒\text{Rate of reaction}=\frac{0.360}{2}=0.180\frac{M}{s}$
Hence the answer is option C

Travis Hicks
Answer : The rate of appearance of $B{r}_{2}$ is, 0.180 M/s
Explanation:
The general rate of reaction is,
$aA+bB⇒cC+dD$
Rate of reaction: It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be:
$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d\left[A\right]}{dt}$
$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d\left[B\right]}{dt}$
$\text{Rate of formation of C}=+\frac{1}{c}\frac{d\left[C\right]}{dt}$
$\text{Rate of formation of D}=+\frac{1}{d}\frac{d\left[D\right]}{dt}$
$Rate=-\frac{1}{a}\frac{d\left[A\right]}{dt}=-\frac{1}{b}\frac{d\left[B\right]}{dt}=+\frac{1}{c}\frac{d\left[C\right]}{dt}=+\frac{1}{d}\frac{d\left[D\right]}{dt}$
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
The given rate of reaction is,
$2HBr\left(g\right)⇒{H}_{2}\left(g\right)+B{r}_{2}\left(g\right)$
The expression for rate of reaction:
Rate of disappearance of $HBr=-\frac{1}{2}\frac{d\left[HBr\right]}{dt}$
Rate of appearance of ${H}_{2}=+\frac{d\left[{H}_{2}\right]}{dt}$
Rate of apperance of $B{r}_{2}=+\frac{d\left[B{r}_{2}\right]}{dt}$
As we are given:
Rate of disappearance of $HBr=0.360\frac{M}{s}$
$+\frac{d\left[B{r}_{2}\right]}{dt}=-\frac{1}{2}\frac{d\left[HBr\right]}{dt}$
$\frac{d\left[B{r}_{2}\right]}{dt}=\frac{1}{2}×0.360\frac{M}{s}$
$\frac{d\left[B{r}_{2}\right]}{dt}=0.180\frac{M}{s}$
Thus, the rate of appearance of