My teacher asked us this question and I don\'t know where to begin, please help

Question

My teacher asked us this question and I don't know where to begin, please help and provide small explanations as you go.

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()

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.

Determine the standard enthalpy of formation for the solid sodium acetate trihydrate using information found in your textbook appendix table.

Explanation / Answer

H3O+(aq) + NaCH3CO23 H2O (s) --- > Na+ (aq) + CH3COOH (aq) + 4H2O(l)

Mass of sodium acetate trihydrate = 1.3584 g

Change in temperature = 0.397 oC

he heat capacity of the calorimeter = 91.0 J/C

standard enthalpy of formation of sodium acetate trihydrate = - 1604 kJ/mol

heat capacity density of aqueous acid solution = 4.144 kJ / mol- mL

molar mass of sodium acetate trihydrate = 135.985 g/mol

enthalpy of reaction = (91 * 0.397 + 4.144 * 0.397 * 100)(1.3584 / 135.985)

= 20.086 kJ/mol

We know that,

enthalpy of reaction = (summation of standard enthalpies of formation of products) - (summation of standard enthalpies of formation of reactants)

20.086 = delta Hf sodium acetate trihydrate + (- 484.5) + 3 * (-28583) - 0 - (- 1604)

delta Hf sodium acetate trihydraet = - 241.924 kJ / mol

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