Define a dynamic equilibrium. The stage in a chemical reaction when the rate of forward reaction equals the rate of reverse reaction. Both forward and reverse reactions take place at the same rate. Use the information in the graph and write down the value of: 6.2.1 x 6.2.2 y 6.2.3 z Calculate the equilibrium constant, Kc, for this hypothetical reaction at room temperature if the volume of the closed container is 3 dm³. At t = 4 minutes, the temperature of the system was increased to 60 °C. Is the REVERSE reaction EXOTHERMIC or ENDOTHERMIC? Explain how you arrived at the answer.

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Define a dynamic equilibrium - English General - NSC Physical Sciences - Question 6 - 2018 - Paper 2

Question 6

Define a dynamic equilibrium. The stage in a chemical reaction when the rate of forward reaction equals the rate of reverse reaction. Both forward and reverse react... show full transcript

Worked Solution & Example Answer:Define a dynamic equilibrium - English General - NSC Physical Sciences - Question 6 - 2018 - Paper 2

Step 1

Define a dynamic equilibrium.

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Answer

A dynamic equilibrium in a chemical reaction occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this stage, both forward and reverse reactions are continuously taking place, ensuring that the concentrations of reactants and products remain constant over time.

Step 2

6.2.1 x

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Answer

From the graph, at t = 2 minutes, the value of x is 8 moles.

Step 3

6.2.2 y

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Answer

At the same time, the value of y is obtained from the graph, which shows 4 moles.

Step 4

6.2.3 z

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Answer

The value of z from the graph at equilibrium is 12 moles.

Step 5

Calculate the equilibrium constant, Kc, for this hypothetical reaction at room temperature.

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Answer

To calculate the equilibrium constant Kc, use the formula:

$K_c = \frac{[C]}{[A]^a[B]^b}$

Given:

Volume = 3 dm³
At equilibrium: z = 12 moles, x = 8 moles, y = 4 moles

Concentrations at equilibrium:

$[C] = \frac{z}{3} = \frac{12}{3} = 4 \text{ mol/dm}^3$
$[A] = \frac{x}{3} = \frac{8}{3} = 2.67 \text{ mol/dm}^3$
$[B] = \frac{y}{3} = \frac{4}{3} = 1.33 \text{ mol/dm}^3$

Substituting the values into the equation:

$K_c = \frac{4}{(2.67)^{3}(1.33)} = 6.75$

Step 6

At t = 4 minutes, the temperature of the system was increased to 60 °C. Is the REVERSE reaction EXOTHERMIC or ENDOTHERMIC?

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Answer

The reverse reaction is ENDOTHERMIC. An increase in temperature typically favors the endothermic reaction, as it absorbs heat from the surroundings. Therefore, as the temperature increases, it shifts the equilibrium position towards the side of the endothermic reaction.

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Define a dynamic equilibrium - English General - NSC Physical Sciences - Question 6 - 2018 - Paper 2

Worked Solution & Example Answer:Define a dynamic equilibrium - English General - NSC Physical Sciences - Question 6 - 2018 - Paper 2

Define a dynamic equilibrium.

6.2.1 x

6.2.2 y

6.2.3 z

Calculate the equilibrium constant, Kc, for this hypothetical reaction at room temperature.

At t = 4 minutes, the temperature of the system was increased to 60 °C. Is the REVERSE reaction EXOTHERMIC or ENDOTHERMIC?

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