Question

WRITING MUST BE CLEAR!!

1. Use the following data to calculate the heat of solution for rubidium iodide in kJ/mol. 9.853 g of RbI was added to a coffee cup calorimeter containing 65.0 mL of water measured at 20.25 ℃. temperature attained after completely dissolving the Rb1 was 17.21 ℃ The lowest Specific heat of water = 4.184 J/g.C: Density of water 0.998 17 g/mL at 20.25℃ Also, is this heat of solution endothermic or exothermic? Heat of Solution Endo- or Exothermic? 2. Today's lab exercise uses a coffee cup calorimeter to measure the heat of reaction (enthalpy of reaction) of several different reactions. We will assume the coffee cup is a perfect insulator, but the reality is the coffee cup minimizes but does not eliminate, heat transfer with the surroundings. Small, but significant, amounts of heat may be either gained from or lost to the environment. How would you expect the heat of solution calculated in question 1 to be impacted by the poor insulating qualities of the coffee cup calorimeter? Specifically, which measured value would be in error? And would the resulting magnitude ofthe ΔH calculated be too large, too small, or not be affected? Explain.

Answer 1

RbI rightarrow 212.37 (M. Wt) Mole (taken) = Wt/M.Wt = 9.853/212.37 RbI = 0.0463 mole Acconding to thermodyamics q downarrow Qty of heat = s downarrow Specific heat m downarrow mass Delta T downarrow difference q = 4.184 times (20.25 - 17.21) times 64.88 Density = m/v m = v times density = 65 times 0.99817 Weight of water m = 64.88105 \r\n q = 4.184 times 64.88 times 3.04 q = 825.23 J for 0.0463 mole of RbI For 1 mole of RbI q = 825.23/0.0463 q = 17,823 J q = 17.823 kJ/mole Since Temperature decreases. It is endothymic dessolution. Delta H = + 17.823 kJ/mole \r\n They will be error since (i) There is loss. (ii) Since it looses to environment. The calculate Delta T will be less than true value. (iii) Since q is directly related with Delta T as below. q = sm DeltaT Delta T