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Karo-lina-s [1.5K]
3 years ago
7

The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is ________ kj. the specific heats of ice, wat

er, and steam are 2.09 j/g−k, 4.18 j/g−k, and 1.84 j/g−k, respectively. for h2o, δ hfus = 6.01kj/mol, and δhvap = 40.67 kj/mol.
Chemistry
1 answer:
liubo4ka [24]3 years ago
3 0

Answer : The enthalpy change for converting 1 mole of ice at -25.0^oC to water at 90^oC is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

(1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]

where,

\Delta H = enthalpy change

m = mass of water = 1mole\times 18g/mole=18g

c_{ice} = specific heat of ice = 2.09 J/gk

c_{water} = specific heat of water = 4.18 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion : 0^oC=273k

T_1 = initial temperature of ice = 0^oC=273k

T_2 = final temperature of ice = -25^oC=273+(-25)=248k

T_3 = initial temperature of water = 0^oC=273k

T_4 = final temperature of water = 90^oC=273+90=363k

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

\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]

\Delta H=7712.106J=7.712KJ     (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at -25.0^oC to water at 90^oC is, 7.712 KJ

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