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Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .?
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Two samples of HI each of 5 gram were taken into two vessels of volume...
**Explanation:**
The extent of dissociation of HI in the two vessels can be explained by Le Chatelier's principle and the principles of chemical equilibrium.
**Le Chatelier's Principle:**
According to Le Chatelier's principle, when a system at equilibrium is subjected to a change in conditions, the system will adjust to minimize the effect of that change and maintain equilibrium.
**Principles of Chemical Equilibrium:**
In a chemical equilibrium, the forward and reverse reactions occur at the same rate. The equilibrium constant (Kc) is a measure of the extent of the reaction. For a dissociation reaction like HI ⇌ H+ + I-, the equilibrium constant can be represented as [H+][I-]/[HI].
Now, let's analyze the two vessels separately to understand the extent of dissociation of HI.
**Vessel 1 (5 liters):**
In this vessel, the volume is smaller, and the concentration of HI is higher compared to vessel 2.
- Initially, when 5 grams of HI is taken into a 5-liter vessel, the concentration of HI is 5/127 = 0.039 M (assuming HI is completely dissociated).
- As the extent of dissociation is equal in both cases, let's assume x moles of HI dissociates.
- At equilibrium, the concentration of HI will decrease by x, and the concentrations of H+ and I- will increase by x.
- Therefore, the equilibrium concentrations can be expressed as [HI] = (0.039 - x) M, [H+] = x M, and [I-] = x M.
- Using the equilibrium constant expression, Kc = [H+][I-]/[HI] = (x * x) / (0.039 - x).
- The value of Kc will remain constant at a given temperature for a specific reaction.
**Vessel 2 (10 liters):**
In this vessel, the volume is larger, and the concentration of HI is lower compared to vessel 1.
- Initially, when 5 grams of HI is taken into a 10-liter vessel, the concentration of HI is 5/254 = 0.019 M (assuming HI is completely dissociated).
- As the extent of dissociation is equal in both cases, let's assume x moles of HI dissociates.
- At equilibrium, the concentration of HI will decrease by x, and the concentrations of H+ and I- will increase by x.
- Therefore, the equilibrium concentrations can be expressed as [HI] = (0.019 - x) M, [H+] = x M, and [I-] = x M.
- Using the equilibrium constant expression, Kc = [H+][I-]/[HI] = (x * x) / (0.019 - x).
- The value of Kc will remain constant at a given temperature for a specific reaction.
**Comparison:**
Since the extent of dissociation of HI is equal in both vessels, it implies that the value of x is the same in both cases. However, the concentrations of HI, H+, and I- are different due to the difference in volume.
- In vessel 1, the higher concentration of HI (0.039 M) leads to a higher concentration of H+ and I- at equilibrium.
- In vessel 2, the lower concentration of HI (0.019 M) leads
The extent of dissociation of HI in the two vessels can be explained by Le Chatelier's principle and the principles of chemical equilibrium.
**Le Chatelier's Principle:**
According to Le Chatelier's principle, when a system at equilibrium is subjected to a change in conditions, the system will adjust to minimize the effect of that change and maintain equilibrium.
**Principles of Chemical Equilibrium:**
In a chemical equilibrium, the forward and reverse reactions occur at the same rate. The equilibrium constant (Kc) is a measure of the extent of the reaction. For a dissociation reaction like HI ⇌ H+ + I-, the equilibrium constant can be represented as [H+][I-]/[HI].
Now, let's analyze the two vessels separately to understand the extent of dissociation of HI.
**Vessel 1 (5 liters):**
In this vessel, the volume is smaller, and the concentration of HI is higher compared to vessel 2.
- Initially, when 5 grams of HI is taken into a 5-liter vessel, the concentration of HI is 5/127 = 0.039 M (assuming HI is completely dissociated).
- As the extent of dissociation is equal in both cases, let's assume x moles of HI dissociates.
- At equilibrium, the concentration of HI will decrease by x, and the concentrations of H+ and I- will increase by x.
- Therefore, the equilibrium concentrations can be expressed as [HI] = (0.039 - x) M, [H+] = x M, and [I-] = x M.
- Using the equilibrium constant expression, Kc = [H+][I-]/[HI] = (x * x) / (0.039 - x).
- The value of Kc will remain constant at a given temperature for a specific reaction.
**Vessel 2 (10 liters):**
In this vessel, the volume is larger, and the concentration of HI is lower compared to vessel 1.
- Initially, when 5 grams of HI is taken into a 10-liter vessel, the concentration of HI is 5/254 = 0.019 M (assuming HI is completely dissociated).
- As the extent of dissociation is equal in both cases, let's assume x moles of HI dissociates.
- At equilibrium, the concentration of HI will decrease by x, and the concentrations of H+ and I- will increase by x.
- Therefore, the equilibrium concentrations can be expressed as [HI] = (0.019 - x) M, [H+] = x M, and [I-] = x M.
- Using the equilibrium constant expression, Kc = [H+][I-]/[HI] = (x * x) / (0.019 - x).
- The value of Kc will remain constant at a given temperature for a specific reaction.
**Comparison:**
Since the extent of dissociation of HI is equal in both vessels, it implies that the value of x is the same in both cases. However, the concentrations of HI, H+, and I- are different due to the difference in volume.
- In vessel 1, the higher concentration of HI (0.039 M) leads to a higher concentration of H+ and I- at equilibrium.
- In vessel 2, the lower concentration of HI (0.019 M) leads
Community Answer
Two samples of HI each of 5 gram were taken into two vessels of volume...
The equilibrium reaction is 2HI(g)⇔H2(g)+I2(g).
In this reaction, the value of Δn is (1+1)−2=0.
The equilibrium of such a reaction is not affected by the change in pressure. Thus the extent of dissociation of HI will be the same in 5 L vessel and 10 L vessel.
In this reaction, the value of Δn is (1+1)−2=0.
The equilibrium of such a reaction is not affected by the change in pressure. Thus the extent of dissociation of HI will be the same in 5 L vessel and 10 L vessel.
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Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .?
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Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .?.
Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two samples of HI each of 5 gram were taken into two vessels of volume 5 and 10 litres each at 27 C . The extent of dissociation of HI is equal in both cases . Explain .?.
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