What is the enthalpy change for producing 1.00 L of acetic acid (d = 1.044 g/mL)

Question

What is the enthalpy change for producing 1.00 L of acetic acid (d = 1.044 g/mL) by this reaction? Assume you mix 100.0 mL of 0 200 M CsOH with 50.0 mL of 0.400 M HCl in a coffee-cup calorimeter. The following reaction occurs: CsOH(aq) + HCl(aq) rightarrow CsCl(aq) + H_2O(l) The temperature of both solutions before mixing was 22.50 degree C, and it rises to 24.28 degree C after the acid-base reaction. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solutions are all 1.00 g/mL and the specific heat capacities of the solutions are 4.2 J/g middot K. You mix 125 mL of 0.250 M CsOH with 50.0 mL of 0.625 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 21.50 degree C before mixing to 24.40 degree C after the reaction. CsOH(aq) + HF (aq) rightarrow CsF(aq) + H_2O(l)

Explanation / Answer

number of moles of CsOH = M*V = 0.200 M * 0.1L = 0.02 mol

number of moles of HCl = M*V = 0.400 M * 0.05 L = 0.02 mol

specific heat capacity, C = 4.2 J/g.K

total volume of solution = 100mL+ 50 mL = 150 mL

Density = 1 g/mL

mass, m = density * volume = 150*1 = 150 g

Heat released, Q = m*C*delta T

= 150*4.2*(24.28 - 22.50)

= 1121.4 J

so 1121.4 J is given out by 0.02 mol of acid-base reaction

enthalphy change per mol = 1121.4/0.02 J/mol = 56070 J/mol

= 56.07 KJ/mol

Answer: 56.07 KJ/mol

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