When 1.3584 g of sodium acetate trihydrate was mixed into 100.0 mL of 0.2000 M H

Question

When 1.3584 g of sodium acetate trihydrate was mixed into 100.0 mL of 0.2000 M HCl (aq) at 25C in a coffee-cup calorimeter, its temperature fell by 0.397C. The reaction occurring is as follows:

H3O+(aq) + NaCH3CO23 H2O (s) Na+ (aq) + CH3COOH (aq) + 4H2O()

a) The heat capacity of the calorimeter is 91.0 J/C. Determine the enthalpy of reaction (in kJ/mol). Describe any assumptions that you made.

b) Determine the standard enthalpy of formation for the solid sodium acetate trihydrate.

I have no clue where to start! Can someone help please! I would greatly appreciate it! :)

Explanation / Answer

H3O+(aq) + NaCH3CO2 3 H2O (s) ----> Na+ (aq) + CH3COOH (aq) + 4H2O(l)

No of moles of NaCH3CO2 3 H2O (s) = wt/mwt

Molar mass of NaCH3CO2*3H2O is 136.0796 g/mol

= 1.3584/136.0796

= 0.0099 mole

No of moles of HCl = M*V = 100/1000*0.2 = 0.02 mole.

heat capacity of the calorimeter C = q/DT

q = heat absorbed = ?

DT = 0.397 C
          
C = 91.0 J/C

q = 91*0.397 = 36.127 joule. = 0.036127 kj

enthalpy of reaction (in kJ/mol) = 0.036127/0.0099 = 3.65 kj/mol

DHrxn = DHfch3cooh + 4*DHfh2o - DHfNaCH3CO2 3 H2O + DHfH3O+

   3.65 = 4*(285.8) + (- 484.2 ) - DHfNaCH3CO2 3 H2O + 0

DHfNaCH3CO2 3 H2O = -1631.05 kj/mol

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