1. 36.7 kj/mol rxn for NaOH (s) = NaOH (aq) 2. -0.4311 kj/mol rxn for HCl(aq)+Na

Question

1. 36.7 kj/mol rxn for NaOH (s) = NaOH (aq)

2. -0.4311 kj/mol rxn for HCl(aq)+NaOH(s)

3. -0.28 kj/mol rex for HCl (aq) +NaOH (aq)

4. -0.277 kj/mol rxn for CH3COOH(aq)+NaOH (aq)

a. find net equation for CH3COOH(aq)+NaOH(aq)

b. explain how could you find the heat of reaction in kj/mol for the ionization of acetid acid using the values you have found in #2 CH3COOH(aq)=H(aq)+CH3COO- (aq)

c. find the numeric value for the heat of reaction in kJ/ mol rxn for the ionization of acetic acid using the values in #2

d. why was a calorimeter constant not calculated

Thank you!

Explanation / Answer

a. Net eq _ CH3COOH (aq) + NaOH (aq) = CH3COO- + Na+ + H2O

b. Heat of reaction = H products - H reactants

c. CH3COOH(aq) = H(aq) + CH3COO- (aq) = -0.28 - (-0.27) =-0.01

d. Experimental measurement of the heat of reaction is known as calorimetry.The heat given out or absorbed in a chemical reaction is measured in a apparatus is called calorimeter. For this reaction, temperature & W is not given to find the calorimeter constant.

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