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Enthalpy change of atomisation of Iodine
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The equation is:
1⁄2 I2(s) ----> I(g)
Why is 1⁄2 I2 solid and not gaseous?
1⁄2 I2(s) ----> I(g)
Why is 1⁄2 I2 solid and not gaseous?
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#2
Because iodine is a solid at room temperature and the enthalpy changes are taken at 298K and 100KPa
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Thank you! So, why does it become gaseous after the atomisation has taken place, do the conditions change during the reaction?
(Original post by Deggs_14)
Because iodine is a solid at room temperature and the enthalpy changes are taken at 298K and 100KPa
Because iodine is a solid at room temperature and the enthalpy changes are taken at 298K and 100KPa
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(Original post by LauraEddy)
Thank you! So, why does it become gaseous after the atomisation has taken place, do the conditions change during the reaction?
Thank you! So, why does it become gaseous after the atomisation has taken place, do the conditions change during the reaction?
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(Original post by Deggs_14)
I believe this is to do with the Maxwell-Boltzman distribution. For a given number of particles in 1/2 I2 (s), some particles will have enough activation energy so the reaction can occur. Enthalpy of atomisation is defined as the amount of energy required for a substance in standard conditions to form one mole of gaseous atoms. As all of the bonds are broken but none are formed, the value is always positive, so energy is required to be inputted into the system, which comes as heat.
I believe this is to do with the Maxwell-Boltzman distribution. For a given number of particles in 1/2 I2 (s), some particles will have enough activation energy so the reaction can occur. Enthalpy of atomisation is defined as the amount of energy required for a substance in standard conditions to form one mole of gaseous atoms. As all of the bonds are broken but none are formed, the value is always positive, so energy is required to be inputted into the system, which comes as heat.
I really just need to get those definitions nailed down :/
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