18 gm of ice is converted into water at 0°C and 1 atm. The entropies of and are 38.2 and 60 J/mol K respectively. The enthalpy change for this conversion is :
Question
18 gm of ice is converted into water at 0°C and 1 atm. The entropies of and are 38.2 and 60 J/mol K respectively. The enthalpy change for this conversion is :
Solution
The enthalpy change for the conversion of ice into water can be calculated using the formula:
ΔH = ΔS * T
where: ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature.
First, we need to calculate the entropy change (ΔS). The entropy of water is greater than the entropy of ice, so the entropy change is positive. We subtract the entropy of ice from the entropy of water:
ΔS = S_water - S_ice = 60 J/mol K - 38.2 J/mol K = 21.8 J/mol K
Next, we convert the temperature from Celsius to Kelvin. The temperature is 0°C, which is equivalent to 273.15 K.
Finally, we substitute these values into the formula to find the enthalpy change:
ΔH = ΔS * T = 21.8 J/mol K * 273.15 K = 5956.47 J/mol
However, the question states that we have 18 gm of ice. The molar mass of water (and ice) is approximately 18 gm/mol, so we have 1 mol of ice. Therefore, the enthalpy change for this conversion is 5956.47 J/mol * 1 mol = 5956.47 J.
Similar Questions
he densities of ice and liquid water at 1 atm (101,325 N/m2) and 0°C (273 K) are 0.917 g/c.c. (0.917 x IO3 kg/m3) and 0.9998 g/c.c. (0.9998 xio3 respectively. The heat of fusion of water is 80 cal/g (334.72 x 10 J/kg). o Calculate the melting point of ice at 0.5 atm (50,662.5 N/m ) and 101 atm (102,338 X IO2 N/m2).
The specific heat of ice is 2.10 kJ/kg °C, the heat of fusion for ice at 0 °C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, and the heat of vaporization of water at 100 °C is 2,256 kJ/kg. What is the final equilibrium temperature when 5.00 grams of ice at −15.0 °C is mixed with 40.0 grams of water at 75.0 °C?
To change solid ice to liquid water ______ Joules are added for every gram. 22601334
How much energy is generated from freezing 2.5 g water?A.2.5 g 1 mol/18.02 g (285.83) kJ/molB.2.5 g 1 mol/18.02 g 40.65 kJ/molC.2.5 g 1 mol/18.02 g 4.186 kJ/molD.2.5 g 1 mol/18.02 g (6.03) kJ/molSUBMITarrow_backPREVIOUS
a. How much heat is needed to convert 1kg of ice at -30oC to steam at 150oC in KJ and Kw-HrB. If the conversion happens in 10 minutes, what is the hp developed?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.