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3.2 moles of HI were heated in a sealed bulb at 444c till the equilibrium state was reached the number of moles of hydrogen present at equilibrium are 0.352 . Find the degree of dissociation of HI at this temperature .?
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3.2 moles of HI were heated in a sealed bulb at 444c till the equilibr...
The degree of dissociation of HI at the given temperature can be calculated as follows:
Let the number of moles of HI that dissociate at equilibrium be x.
Since the reaction is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.
According to the given information, the number of moles of hydrogen present at equilibrium is 0.352. Since 1 mole of HI yields 1 mole of hydrogen when dissociated, the number of moles of HI that dissociate at equilibrium is also 0.352.
Thus, the degree of dissociation of HI at the given temperature is 0.352 / 3.2 = 0.11.
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3.2 moles of HI were heated in a sealed bulb at 444c till the equilibr...
Given:
- Initial moles of HI = 3.2 moles
- Temperature = 444°C
- Moles of hydrogen at equilibrium = 0.352 moles

To find:
- Degree of dissociation of HI at this temperature

Explanation:
The degree of dissociation of a substance can be defined as the fraction of the substance that has dissociated into its constituent elements at equilibrium.

To find the degree of dissociation of HI at this temperature, we need to use the balanced chemical equation for the dissociation of HI:

2HI ⇌ H2 + I2

Let's assume that the degree of dissociation of HI is α. This means that α moles of HI dissociate to form α moles of H2 and α moles of I2.

Step 1: Calculating moles of HI that dissociate:
Since α moles of HI dissociate to form α moles of H2 and α moles of I2, the moles of HI that dissociate can be calculated as follows:

Moles of HI dissociated = α * moles of HI
= α * 3.2 moles

Step 2: Calculating moles of H2 and I2 formed:
Since 2 moles of HI dissociate to form 1 mole of H2 and 1 mole of I2, the moles of H2 and I2 formed can be calculated as follows:

Moles of H2 formed = 0.5 * Moles of HI dissociated
= 0.5 * α * 3.2 moles

Moles of I2 formed = 0.5 * Moles of HI dissociated
= 0.5 * α * 3.2 moles

Step 3: Calculating moles of HI remaining:
The moles of HI remaining after dissociation can be calculated as follows:

Moles of HI remaining = Initial moles of HI - Moles of HI dissociated
= 3.2 moles - α * 3.2 moles
= 3.2(1 - α) moles

Step 4: Writing the equilibrium expression:
The equilibrium expression for the dissociation of HI can be written as:

Kc = ([H2] * [I2]) / [HI]^2

Since the moles of H2 formed and moles of I2 formed are equal to α * 3.2 moles, and the moles of HI remaining is equal to 3.2(1 - α) moles, the equilibrium expression becomes:

Kc = (α * 3.2 * α * 3.2) / (3.2(1 - α))^2

Step 5: Calculating the value of Kc:
Since we know the value of Kc at this temperature is equal to 0.352, we can use this information to calculate the value of α.

0.352 = (α * 3.2 * α * 3.2) / (3.2(1 - α))^2

Solving this equation will give us the value of α, which represents the degree of dissociation of HI at this temperature.

Conclusion
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3.2 moles of HI were heated in a sealed bulb at 444c till the equilibrium state was reached the number of moles of hydrogen present at equilibrium are 0.352 . Find the degree of dissociation of HI at this temperature .?
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