For the evaporation of water, H₂O(l) → H₂O(g), at 100°C, ∆H = +9.72 kcal/mol (+40.7 kJ/mol).
b. How many kilojoules are released when 10.0 g of H₂O(g) is condensed?

Question

For the evaporation of water, H₂O(l) → H₂O(g), at 100°C, ∆H = +9.72 kcal/mol (+40.7 kJ/mol).

b. How many kilojoules are released when 10.0 g of H₂O(g) is condensed?

Accepted Answer

Welcome back everyone. For the sublimation of dry ice, we have carbon dioxide solid which forms carbon dioxide gas were given an entropy change of positive 25 point okay, joules per mole. And we need to calculate the energy and killing tools required for the sublimation of 8.5 g of dry ice. Begin by recalling that sublimation describes our conversion of solid form of our matter to its gaseous form. And delta H represents our entropy change or heat exchange in our system being our reaction, Which we're told is equal to positive 25.0 killing joules per mole. Let's recall that when we have a positive value for entropy, this tells us that our reaction is going to be endo thermic, which we should recall therefore tells us that our reaction being our system absorbs energy or in other words, energy is required. So we need to figure out how much energy is required. And in this case, energy can be considered heat. And so we're going to recall that our term for heat represents Q. So we want to solve for Q And beginning with our mass of our sample of solid ice we have of dry ice. We have 8.5 g of dry ice, CO2 solid and recall that our unit for heat is expressed in killing jewels as the prompt states. So we want to cancel out our unit of g. Let's begin by recalling our molar mass for carbon dioxide. Where in the denominator from a periodic table, we see a mass of 44.009g of CO2 with an equivalent of one mole of CO2 in our numerator. So using the smaller mass, we're now able to cancel out grams of C 02 leaving us with molds of C 02. But again, we need killer jewels as our final unit. And this is where we're going to use our given entropy change as a conversion factor which we multiply by. And so what we would interpret is that we have per mole of carbon dioxide reacted in our denominator, we have the release of positive 25.0 kill jewels of heat or rather the absorption. In this case, since again, the reaction is endo thermic. And so multiplying by this conversion factor, we're now able to cancel out our units of moles of co two leaving us with killing jewels as our final unit for heat And in our calculators, this will result in a value of 4.828 and sorry, four kill jewels. Of heat that is required for the sublimation of our dry ice carbon dioxide solid. So we want to round this to our least precise measurement as two significant figures based on our given mass of 8.5 g surrounding to two sig figs. We would come up with our final answer as positive 4.8 kg joules of heat that is absorbed or required. And this would be our final answer corresponding to choice a in the multiple choice as our energy required for the sublimation of our 8.5 g of dry ice. So I hope that this made sense and let us know if you have any questions.

For the evaporation of water, H₂O(l) → H₂O(g), at 100°C, ∆H = +9.72 kcal/mol (+40.7 kJ/mol).
b. How many kilojoules are released when 10.0 g of H₂O(g) is condensed?

Verified video answer for a similar problem:

Key Concepts

Enthalpy of Vaporization

Molar Mass of Water

Energy Conversion

For the evaporation of water, H2O(l) → H2O(g), at 100°C, ∆H = +9.72 kcal/mol (+40.7 kJ/mol). b. How many kilojoules are released when 10.0 g of H2O(g) is condensed?

Verified video answer for a similar problem:

Key Concepts

Enthalpy of Vaporization

Molar Mass of Water

Energy Conversion

For the evaporation of water, H₂O(l) → H₂O(g), at 100°C, ∆H = +9.72 kcal/mol (+40.7 kJ/mol).
b. How many kilojoules are released when 10.0 g of H₂O(g) is condensed?