Question

The salt potassium bromide dissolves in water according to the reaction:

KBr(s) K+(aq) + Br-(aq)

(a) Calculate the standard enthalpy change ΔH° for this reaction, using the following data:

KBr(s) = -393.8 kJ mol-1

K+(aq) = -252.4 kJ mol-1

Br-(aq) = -121.6 kJ mol-1

________kJ

(b) Suppose 63.5 g of KBr is dissolved in 0.186 L of water at 24.3 °C. Calculate the temperature reached by the solution, assuming it to be an ideal solution with a heat capacity close to that of 186 g of pure water (specific heat = 4.18 J g-1 °C-1).

_________°C

(c) Heats of reaction find practical application in hot packs or cold packs. Would this dissolution reaction be appropriate for the preparation of a hot pack or a cold pack?

Hot or cold pack? _______

Answer 1

(a) KBr(s) ----> K+(aq) + Br-(aq) :  ΔH° = ?

ΔH° =  ΔH°_f products -  ΔH° _f reactants

=  [ΔH°_f K+ (aq)) + ΔH°_f Br-(aq) ] - [ ΔH°_f KBr(s)]

= [(-252.4) + (-121.6)]-[-393.8] kJ/mol

= 19.8 kJ/mol

(b) Molar mass of KBr is 119 g/mol

From the aboveequation ,

1 mole = 119 g of KBr releases 19.8 kJ of heat

63.5 g of KBr releases ( 19.8x63.5) / 119 = 10.6 kJ of heat

Heat absorbed by water , q = 10.6 kJ =10.6x10³ J= mcdt

where m = mass of water = volume x density

= 0.186 L x ( 1000 mL/L) x 1.0 g/mL

= 186 g

c = specific heat capacity of water = 4.18J g-1 °C-1

dt = change in temperature = final - initial = t -24.3

Plug the values we get  

dt = q / ( mc) = 13.6

t - 24.3 = 13.6

t = 37.9 oC