# Calculate the enthalpy change for the process in which 27.5 g of water is converted from liquid...

## Question:

Calculate the enthalpy change for the process in which 27.5 g of water is converted from liquid at 6.8 degrees Celsius to vapor at 25.0 degrees Celsius. For water, {eq}\Delta H_{\large \rm vap} = 44.0 \ \rm kJ/mol {/eq} at {eq}\rm 25.0^{\circ} C {/eq} and {eq}C_s = 4.18 \ \rm J/(g \cdot {^{\circ}C}) {/eq} for {eq}\rm H_2O(l) {/eq}.

## Heating Curve:

Heating curve shows the temperature change of a substance vs time (or heat added). The slopped region refers to temperature change of a single phase while the flat region is when two phases are in equilibrium, or a phase change. Energy will go towards phase change at phase transition temperature before temperature change.

## Answer and Explanation:

First we will convert the mass of water to moles

$$27.5 \ g \times \dfrac {1 \ mole}{18.02 \ g} =1.526 \ mole $$

The temperature change of the water is from 6.8 to 25.0 {eq}^oC {/eq} or

$$\Delta T = 25.0 - 6.8 = 18.2 ^oC $$

The heat involved in heating the water (q) is then calculated using the specific heat (C) as below

$$q = mC \Delta T\\ q = (27.5 \ g)(4.18 \dfrac {J}{g \cdot ^oC}) (18.2 ^oC)\\ q = 2092 \ J = 2.092 \ kJ $$

The phase change from water to vapor is then calculated using the given heat of vaporization

$$q = (mole \ H_2O) \Delta H_{vap}\\ q = (1.526 \ mole)(44.0 \ kJ/mole)\\ q = 67.15 \ kJ $$

The total enthalpy change is then calculated by adding the total heat transferred and then divide by the mole of water

$$\Delta H= \dfrac {q_{total}}{mole}\\ \Delta H= \dfrac {2.092 \ kJ + 67.15 \ kJ}{1.526 \ mole}\\ \Delta H= \dfrac {69.2397 \ kJ}{1.526 \ mole}\\ \Delta H= \boxed {45.4 \ kJ/mole} $$

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