Question

(1):

When a 1.48-g sample of solid ammonium nitrate dissolves in 56.4 g of water in a coffee-cup calorimeter (see above figure) the temperature falls from 22.00 ^oC to 20.09 ^oC. Calculate H in kJ/mol NH₄NO₃ for the solution process. NH₄NO₃(s) NH₄⁺(aq) + NO₃^-(aq)
The specific heat of water is 4.18 J/g-K.
H_solution = __kJ/mol NH₄NO₃.

(2):

The reaction S₂O₈^2-(aq) + 3 I^-(aq) 2 SO₄^2-(aq) + I₃^-(aq) was studied at a certain temperature with the following results:

Experiment	[S2O82-(aq)] (M)	[I-(aq)] (M)	Rate (M/s)
1	0.0309	0.0309	5.99e-06
2	0.0309	0.0618	1.20e-05
3	0.0618	0.0309	1.20e-05
4	0.0618	0.0618	2.39e-05

(a) What is the rate law for this reaction?

Rate = k [S₂O₈^2-(aq)] [I^-(aq)]Rate = k [S₂O₈^2-(aq)]² [I^-(aq)]    Rate = k [S₂O₈^2-(aq)] [I^-(aq)]²Rate = k [S₂O₈^2-(aq)]² [I^-(aq)]²Rate = k [S₂O₈^2-(aq)] [I^-(aq)]³Rate = k [S₂O₈^2-(aq)]⁴ [I^-(aq)]

(b) What is the value of the rate constant?

(c) What is the reaction rate when the concentration of S₂O₈^2-(aq) is 0.0602 M and that of I^-(aq) is 0.0655 M if the temperature is the same as that used to obtain the data shown above?
__ M/s

Answer 1

1) Mass of solution = (1.48 g + 56.4 g) = 57.88 g Specificheat of solution, s = 4.184 J/g °C Change in temperature AT = 20.09°C-22.00°C =-1.91°C Amount of heat absorbed in the reaction is equal to the heat relased by the calorimeter. Im = som = -mshAT - - 57.88gx4.184 *(-1.91c)) = +462.5 J = 0.0185 mol NH, NO: 1 mol NH NO3 No of moles of NH.NO3 = 1.48 g x- 80.05 g +462.5 J Enthalpy change, A7 solution = E, a solution 0.0185 mol = +25000 J/mol = +25.0 kJ/mol