Assume you mix 100.0mL of .200M CsOH with 50.0 mL of .400 M HCl ina coffee-cup calorimeter. The following reaction occurs:
CsOH (aq)+ HCl (aq) CsCl(aq) + H2O (l)
The temperature of both solution before mixing was 22.5 *C and it rises to 24.28 *C after the acid-base rxn. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solution are 1.00 g/ml and specific heat of solution is 4.2J/g*k.
PS - The correct answer is Delta H = -56 kJ per mol CsOH
CsOH (aq)+ HCl (aq) CsCl(aq) + H2O (l)
The temperature of both solution before mixing was 22.5 *C and it rises to 24.28 *C after the acid-base rxn. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solution are 1.00 g/ml and specific heat of solution is 4.2J/g*k.
PS - The correct answer is Delta H = -56 kJ per mol CsOH
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first, figure out how many moles of each solution is added
CsOH: .2*.1 = .02mol, HCl: .4*.05 = .02mol
since 150ml is used, theres 150g, so use the formula q=mcT,
q=150*4.2*(24.28-22.5) = 1121.41J per .02mol = 56kJ per mol
PS - idk y delta H is negative... the temperature rises so delta H should be positive?
CsOH: .2*.1 = .02mol, HCl: .4*.05 = .02mol
since 150ml is used, theres 150g, so use the formula q=mcT,
q=150*4.2*(24.28-22.5) = 1121.41J per .02mol = 56kJ per mol
PS - idk y delta H is negative... the temperature rises so delta H should be positive?