
Given that the rate of disappearance of bromine is $$ - 3.5 \times {10^{ - 4}}M/s$$ for the following reaction:
$${H_2}(g) + B{r_2}(g) \to 2HBr(aq)$$
What is the rate of formation for HBr?
A. $$ - 7.0 \times {10^{ - 4}}M/s$$
B. $$ - 3.5 \times {10^{ - 4}}M/s$$
C. $$3.5 \times {10^{ - 4}}M/s$$
D. $$7.0 \times {10^{ - 4}}M/s$$
Answer
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Hint: The rate of reaction is defined as the speed of the reaction during which the formation of products takes place from the reaction. The rate of reaction depends on the type and the nature of reaction. It also depends upon the physical state of reactants, the amount of reaction and the complex nature of reaction.
The formula for rate of reaction
$$reactant \to product$$
Rate of reaction=$$\dfrac{{ - d[reac\tan t]}}{{dt}} = \dfrac{{d[product]}}{{dt}}$$
The negative sign here is used to show the decrease in the concentration of the particular reactant.
Complete step by step answer:
The equation for HBr given is
$${H_2}(g) + B{r_2}(g) \to 2HBr(aq)$$
So for this the formula for rate of reaction will be
Rate of reaction $$ = \dfrac{{ - d[{H_2}]}}{{dt}} = \dfrac{{ - d[B{r_2}]}}{{dt}} = \dfrac{1}{2}\dfrac{{d[HBr]}}{{dt}} = 3.5 \times {10^{ - 4}}$$
Now comparing HBr and bromine we get,
$$\dfrac{{ - d[B{r_2}]}}{{dt}} = \dfrac{1}{2}\dfrac{{d[HBr]}}{{dt}}$$ $$ = + 3.5 \times {10^{ - 4}} \times 2$$
Now substituting the value of rate of disappearance of bromine
$$ - ( - 3.5 \times {10^{ - 4}}) = \dfrac{1}{2}\dfrac{{d[HBr]}}{{dt}}$$
$$\dfrac{{d[HBr]}}{{dt}}$$$$ = + 3.5 \times {10^{ - 4}} \times 2$$
$$\dfrac{{d[HBr]}}{{dt}}$$=$$7.0 \times {10^{ - 4}}M/s$$
So, the correct answer is Option D.
Note: The rate of the reaction increases with increase in the concentration of the reactant. Pressure increases the concentration of reactant so the rate of reaction increases. If the solvent is ionic then the rate of reaction will increase. Increase in the intensity of light, so the rate of reaction increases.
Activation energy is defined as the minimum amount of energy which is required for the activation of atoms or molecules so that it can undergo chemical transformation. There are some conditions of collisions so that the product can be formed and they are that collisions should be effective, a minimum amount of energy should be possessed by all the molecules to form a product.
The formula for rate of reaction
$$reactant \to product$$
Rate of reaction=$$\dfrac{{ - d[reac\tan t]}}{{dt}} = \dfrac{{d[product]}}{{dt}}$$
The negative sign here is used to show the decrease in the concentration of the particular reactant.
Complete step by step answer:
The equation for HBr given is
$${H_2}(g) + B{r_2}(g) \to 2HBr(aq)$$
So for this the formula for rate of reaction will be
Rate of reaction $$ = \dfrac{{ - d[{H_2}]}}{{dt}} = \dfrac{{ - d[B{r_2}]}}{{dt}} = \dfrac{1}{2}\dfrac{{d[HBr]}}{{dt}} = 3.5 \times {10^{ - 4}}$$
Now comparing HBr and bromine we get,
$$\dfrac{{ - d[B{r_2}]}}{{dt}} = \dfrac{1}{2}\dfrac{{d[HBr]}}{{dt}}$$ $$ = + 3.5 \times {10^{ - 4}} \times 2$$
Now substituting the value of rate of disappearance of bromine
$$ - ( - 3.5 \times {10^{ - 4}}) = \dfrac{1}{2}\dfrac{{d[HBr]}}{{dt}}$$
$$\dfrac{{d[HBr]}}{{dt}}$$$$ = + 3.5 \times {10^{ - 4}} \times 2$$
$$\dfrac{{d[HBr]}}{{dt}}$$=$$7.0 \times {10^{ - 4}}M/s$$
So, the correct answer is Option D.
Note: The rate of the reaction increases with increase in the concentration of the reactant. Pressure increases the concentration of reactant so the rate of reaction increases. If the solvent is ionic then the rate of reaction will increase. Increase in the intensity of light, so the rate of reaction increases.
Activation energy is defined as the minimum amount of energy which is required for the activation of atoms or molecules so that it can undergo chemical transformation. There are some conditions of collisions so that the product can be formed and they are that collisions should be effective, a minimum amount of energy should be possessed by all the molecules to form a product.
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