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ohaa [14]
3 years ago
8

At equilibrium the reactant and product concentrations are constant because a change in one direction is balanced by a change in

the other as the forward and reverse rates become equal:At equilibrium: rate_twd=rate_rev When a chemical system is at equilibrium,a. he concentrations of the reactants and products have reached constant values.b. the concentrations of the reactants are equal to the concentrations of the products.c. the forward and reverse reactions have stopped.d. the reaction quotient, Q, has reached a maximum.e. None of the choices.
Chemistry
1 answer:
Anna [14]3 years ago
5 0

Answer: a. The concentrations of the reactants and products have reached constant values

Explanation:

The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.  For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equal to rate of the backward reaction.

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time and are constant.

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For a equilibrium reaction,

A\rightleftharpoons B

K_{eq}=\frac{[B]}{[A]}

Thus the correct answer is the concentrations of the reactants and products have reached constant values.

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a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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