The following reaction occurred in a calorimeter: Ca(OH)_2(s) + 2 HCI(aq) righta
ID: 527789 • Letter: T
Question
The following reaction occurred in a calorimeter: Ca(OH)_2(s) + 2 HCI(aq) rightarrowCaCI_2(aq) + 2 H_2O(l) The complete reaction of 1.60 g of solid Ca(OH)_2 with 100 mL of 1 M HCI produced a temperature increase from 23.9 degree C to 30.3 degree C. Assume that the density of the HCI solution is 1.00 g/mL and the specific heat of the final solution is 4.18 J/g degree C. Calculate the delta H of the reaction in kJ/mol Ca(OH)_2 Use your answer to (a) and the data in Appendix C to calculate delta H^degree_f of CaCl_2(aq).Explanation / Answer
(a) Heat change. q = m * s * (t2 - t1)
q = 100 * 4.18 * (30.3 - 23.9)
q = 2675.2 J
THis much of heat is produced by 1.60 g. of Ca(OH)2
Moles of Ca(OH)2 = mass / molar mass = 1.60 / 74 = 0.0225 mol
Therefore, detaH = q /n = 2675.2 / 0.0225 = 118898 J/mol = 119. kJ/mol
(b)
detaH0rxn = deltaH0f(CaCl2) + 2 deltaH0f(H2O) - deltaH0f(Ca(OH)2) - 2 deltaH0f(HCl)
- 119. = deltaH0f(CaCl2) + 2 ( - 285.8 ) - ( - 986.6 ) - 2 ( - 74.8 )
deltaH0f(CaCl2, aq.) = - 683.6 kJ
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