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A student dissolved 10.5 g of NH 4 Cl in 40 mL of water. The temperature of the

ID: 1066591 • Letter: A

Question

A student dissolved 10.5 g of NH4Cl in 40 mL of water.  The temperature of the water changed from 22.5oC to 5.1oC.  Calculate the enthalpy of hydration of ammonium chloride in J/mol.

1.  A student collected the data below while performing a kinetics experiment.

Trial

Volume of 2.0 M Reactant A

Volume of 0.5 M Reactant B

Volume of H2O

Time for B to react

1

10 mL

10 mL

30 mL

135 sec

2

20 mL

10 mL

20 mL

69 sec

3

10 mL

20 mL

20 mL

34 sec

           

Use the above information to complete the following table (enter all answers with 3 significant figures)

Trial

concentration of A

concentration of B

Rate for B

1

2

3

Using the data from the previous problem, determine the order with respect to Substance A. Using the data from Problem 1, determine the order with respect to Substance B. Using the information from Trial 1 and the experimentally determined rate law, determine the value of the rate constant, k for this experiment. Report your answer with 3 significant figures.

Please answer all the questions and with the three significant digits after decimals. Please.

Trial

Volume of 2.0 M Reactant A

Volume of 0.5 M Reactant B

Volume of H2O

Time for B to react

1

10 mL

10 mL

30 mL

135 sec

2

20 mL

10 mL

20 mL

69 sec

3

10 mL

20 mL

20 mL

34 sec

Explanation / Answer

Question : A student dissolved 10.5 g of NH4Cl in 40 mL of water.  The temperature of the water changed from 22.5oC to 5.1oC.  Calculate the enthalpy of hydration of ammonium chloride in J/mol

solution:

dT    =   5.1oC - 22.5oC = -17.4 oC

volume of water = 40 mL

density of water = 1 g / mL

mass = density x volume

           = 1 x 40

           = 40 g

heat = q = m Cp dT

q   = 40 x 4.184 x (-17.4)

q = -2912.1 J

moles of NH4Cl = 10.5 / 53.5 = 0.196

enthalpy of hydration of ammonium chloride   =   - q / n

                                                                        = - (-2912.1_ / 0.196

                                                                         =   14838 J / mol

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