Question

What is the delta H when 72.0 grams H2O condenses at 100.00C?

Here are some constants that you MAY need.

specific heats heat of fusion heat of vaporization

H2O(s) = 2.1 J/g0C 6.01 kJ/mole 40.7 kJ/mole

H2O(L) = 4.18 J/g0C

H2O(g) = 1.7 J/g0C



2930 kJ


163 kJ


-163 kJ


-2930 kJ

Answer 1

Answer: The value of
`\Delta H` is -163 kJZ

Step-by-step explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

We are given:

Given mass of water = 72.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

`\text{Moles of water}=(72.0g)/(18g/mol)\\\\\text{Moles of water}=4mol`

Calculating the heat released for the condensation process:

`\Delta H=n* \Delta H_((vap))` ......(2)

where,

`\Delta H` = amount of heat released

n = number of moles of water = 4 moles

`\Delta H_((vap))` = specific heat of vaporization = -40.7 kJ/mol

Negative sign represents the amount of heat released.

Putting values in equation 2:

`\Delta H=4mol* (-40.7kJ/mol)=-163kJ`

Hence, the value of
`\Delta H` is -163 kJ