The salt potassium bromide dissolves in water according to the reaction: KBr(s)
ID: 540496 • Letter: T
Question
The salt potassium bromide dissolves in water according to the reaction:
KBr(s) K+(aq) + Br-(aq)
(a) Calculate the standard enthalpy change H° 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? _______
Explanation / Answer
(a) KBr(s) ----> K+(aq) + Br-(aq) : H° = ?
H° = H°f products - H° f reactants
= [H°f K+ (aq)) + H°f Br-(aq) ] - [ H°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