A.) Calculate the enthalpy change, H , for the process in which 37.9 g of water

Question

A.) Calculate the enthalpy change, H, for the process in which 37.9 g of water is converted from liquid at 4.6 C to vapor at 25.0 C .

For water, Hvap = 44.0 kJ/mol at 25.0 C and Cs = 4.18 J/(gC) for H2O(l).

B.)How many grams of ice at -29.4 C can be completely converted to liquid at 20.7 C if the available heat for this process is 5.90×103 kJ ?

For ice, use a specific heat of 2.01 J/(gC) and Hfus=6.01kJ/mol.

C.)In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 C. If 4.20 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution Hsoln of CaCl2 is 82.8 kJ/mol.

D.)

Calculate the standard enthalpy change for the reaction

2A+B2C+2D

Use the following data:

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

d) 2A + B <--> 2C + 2D

delta Hrkn = 2*(delta Hf D + delta Hf C) - 2*delta Hf A - delta Hf B = 399 J/K

c) Heat evolved = mass of solution*specific heat capacity of the solution*change in temperature

or, 82800 = 104.2*4.184*T ; where T = change in temperature

ot, T = 189.92 0C

Thus, final temperature = 189.92 + 23 = 212.92 0C

b) Let x' g of ice can be converetd using the available heat

Now, heat required to bring ice to 00C = mass*specific heat*change in temperature = x*2.01*29.4 = 59.094*x

heat required to convert ice to water at 0 0C = x*333.89

Now, heat required to bring water at 0 0C to 20.7 0C = mass*specific heat capacity of ater*change in temperature = x*4.184*20.7 = 86.609*x

Thus, total heat required = x*[86.609 + 333.89 + 59.094] = 5900000

or, x = 12302 g = 12.302 kg

a) Heat required to bring water to 100 0C = mass*specific heat*change in temperature = 37.9*4.184*(100-4.6) = 15128 J

Heat required to convert water at 100 0C to steam = mass*heat of vaporisation = 37.9*44000/18 = 92644.45 J

Thus, total heat required = 107772.45 J = 107.772 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.