Question

Calculate the enthalpy change associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 110.0 °C. The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For , ΔHfus = 6.01 kJ/mol and ΔHvap = 40.67 kJ/mol.

Answer 1

Answer : The enthalpy change is, 104.5327 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-277K)\rightarrow H_2O(s)(273K)\\\\(2):H_2O(s)(273K)\rightarrow H_2O(l)(273K)\\\\(3):H_2O(l)(273K)\rightarrow H_2O(l)(373K)\\\\(4):H_2O(l)(373K)\rightarrow H_2O(g)(373K)\\\\(5):H_2O(g)(373K)\rightarrow H_2O(g)(383K)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 25 g

$c_{p,s}$ = specific heat of solid water = 2.09 J/gk

$c_{p,l}$ = specific heat of liquid water = 4.18 J/gk

$c_{p,g}$ = specific heat of liquid water = 1.84 J/gk

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{25g}{18g/mole}=1.39mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[25g\times 4.18J/gK\times (273-277)k]+1.39mole\times 6010J/mole+[25g\times 2.09J/gK\times (373-273)k]+1.39mole\times 40670J/mole+[25g\times 1.84J/gK\times (383-373)k]$

$\Delta H=104532.7J=104.5327KJ$     (1 KJ = 1000 J)

Therefore, the enthalpy change is, 104.5327 KJ