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ale4655 [162]
3 years ago
15

A reaction is started with both reactants and products and proceeds in the reverse direction to reach equilibrium. Which of the

following statements best describes why the reaction proceeds in the reverse direction?
A) The initial rate of the reverse reaction is greater than the initial rate of the forward reaction.
B) The rate constant of the reverse reaction is greater than the rate constant of the forward reaction.
C) The activation energy of the reverse reaction is less than the activation energy of the forward reaction.
D) The reactants are lower in energy than the products.
Chemistry
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

Option A

Explanation:

A) Yes. The reaction reaches equilibrium when the rate of reaction of the reverse reaction is equal to the rate of the forward reaction , then the only cause for the reverse reaction to be favoured is that the initial rate of the reverse was greater than the forward one.

B) No. The rate constant of the reverse reaction can be greater than the forward one  but the rate also depends on concentrations, thus a reverse reaction with greater rate constant can result in the net reaction proceeding in the forward reaction, the reverse reaction or be at equilibrium depending on the concentrations or reactants and products

C) No. A lower activation energy means a higher rate constant , but a higher rate constant does not mean that the net reaction will proceed to the reactants ( see point B)

D) No. The energy changes determine conditions under thermodynamic equilibrium and therefore the net direction of the reaction will depend on the temperature and concentrations of reactants and products with respect to the equilibrium conditions.

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
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Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

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First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

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