Question

Chem 1 problem. Please answer all with steps. Thank you so much

2. When 26.6 g of potassium chloride dissolves in 300.0 g of water, the temperature drops from 26.1°C to 11.7°C. The C of the calorimeter is 22.0 J/oC. cal (a) Write an equation for solid potassium chloride dissociating into ions when it dis- solves in water. (b) Is the reaction endothermic or exothermic? (c) What is the sign of AH? (d) What is the value of "q" for KCl in this reaction? (e) What is the value of AH for KCI in this reaction?

Answer 1

a) Equation : KCl(aq) Water > K+ (aq) + C1-(aq) b) Since the temperature is decreased the reaction is an endothermic reaction c) the sign of AH is +ve d) Mass of water, m= 300g Specificheat of water ,s = 4.184 J/g.°C Change in temperature AT = 11.7°C-26.1°C=-14.4°C Amount of heat released by water, water = mSAT g'c = 300 gx4.184– ***(-14.4°C) =-18075 J = -18.075 kJ Change in temperature AT = 11.7°C-26.1°C=-14.4°C Heat capacity of calorimeter ,s = 22.05/°C= 0.0220 kJ/°C Amount of heat absorbed by the calorimeter, cal = CAT = 0.0220 d) (-14.4°C) = -0.3168 kJ Amount of heat gained in the reaction is equal to the heat released by system. system =- ( 9ca1 +4water) =-(-0.3168 kJ+(-18.075 kJ) ) = +18.3918 kJ e) No of moles of Ka = 26.6 gx- 1 mol KC1 – 0.3568 mol +18.3918 kJ Heat of reaction, AH..= - 0.3568 mol = +51.5 kJ/mol