I'm reallt stuck!!!!!!!! with a substance initially at 20.00 degree C, 5.00 g of solid sodium nitrate was dissolved in 100.0 g

Question

With a substance initially at 20.00 degree C, 5.00 g of solid sodium nitrate was dissolved in 100.0 g of water in a calorimeter, and the temperature of the resultant solution was 14.80 degree C. The specific heat of the aqueous solution was 4.20 J/g C, and the heat capacity of the calorimeter was 40.0 J/ C.

a. write a chemical equation for the dissolution of solid sodium nitrate in water
( is it 2NaNO3 (aq) + H2O --> 2HNO3 + Na2O or is it totally a different thing?)

b. was the dissolution of sodium nitrate in water endothermic, exothermic, neither, or cannot be determined? justify your choice
(is it endothermic? b/c your adding heat to dissolve it ? )

c. calculate the energy change per mole of sodium nitrate for dissolving solid sodium nitrate in water
i have no clue.... 

Answer 1

Nerdy,

 

a. For the chemical equation you must realize that this is a "simple" dissolution of a solid ionic compound into its aqueous ions.

 

NaNO₃(s) -> Na⁺(aq) + NO₃^-(aq)

 

b. Since the temperature of the system went down, 25.00 C to 14.80 C, the process must have been endothermic. The system absorbed energy from its surroundings.

 

c. For this question you must first calculate the energy change for the system and second calculate the energy change per mole of sodium nitrate.

 

First the Energy Change

 

Remember that system (sys) and the surroundings (surr) exchange energy and that the surroundings are the solution (soln) and the calorimeter (cal) and that we know how the solution and the calorimeter exchange energy with changes in temperature - that is we know their heat capacities. So let's set up the problem.

 

q_sys = - q_surr

 

q_surr = q_soln + q_cal

 

qsoln = - mcΔT

 

Where m is the mass of the surroundings or solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature of the surroundings or solution.

q_cal = CΔT 

 

Where C is the heat capacity of the calorimeter and ΔT is the change in temperature of the calorimeter.

 

Note: We assume that the solution undergoes the same change in temperature as the calorimeter.

 

Organize your information:

 

m = 5.00 g NaNO₃ + 100.0 g H₂O = 105.0 g total solution

 

c = 4.20 J/g•C

 

ΔT = Final Temperature - Initial Temperature = T_f - T_i = 14.80 C - 20.00 C = - 5.20 C

 

C = 40.0 J/C

 

So we have

 

q_sys = - ((105.0 g x 4.20 J/g•C x - 5.20 C) + (40.0 J/C x - 5.20 C))

 

    = - ( - 2290 J - 208 J)

 

    = + 2500 J

 

Second the Energy Change per Mole

 

First calculate the number of moles of sodium nitrate NaNO₃.

 

5.0 g NaNO₃ x 1 mol NaNO₃/85.0 g NaNO₃ = 0.0588 mol NaNO₃

 

Finally divide the energy change by the number of moles and convert to kilojoules.

 

2500 J/0.0588 mol NaNO₃ = 42,500 J/mol NaNO₃ or 42.5 kJ/mol NaNO₃

 

Note that this answer is positive which is consistent with the prediction that the reaction was endothermic.

 

Please keep in mind that I am available for online tutoring.

 

Thanks

 

Thomas