You must show all work. 1.(4) The DH rxn of any strong acid with any strong base

Question

You must show all work.

1.(4) The DHrxn of any strong acid with any strong base is somewhere between -53.5 and -56.5 kJ/mole, depending on your source. Assume the DHrxn is -53.5 kJ/mole, the specific heat of any dilute water solution is 4.18 J/g Dt, and the density of any dilute water solution is 1.0 g/mL. In a “coffee cup calorimeter” (open system), 3.00 g of sodium hydroxide was added to 50.0 mL of water. Then it was mixed with 75.0 mL of 1.00 M HCl. The initial temperature of both solutions was 21.50C. The final temperature of the mixture was 26.80C. Calculate the joules of heat released in this reaction.

1a. Using your answer to question 1, calculate the DHrxn for your data in kJ/mole of NaOH and your percent error.

Explanation / Answer

total volume of solution = 50 + 75 = 125 ml

density of solution = density of water = 1 g/ml

mass of solution = density x volume = 1 x 125 = 125g

Cp of solution = 4.184 J / oC g

dT = T2 -T1 = 26.80-21.50 = 5.30 oC

hear released = q = m Cp dT

q = 125 x 4.184 x 5.30 = 2771.9 J

hear released   = 2771.9 J

moles of NaOH = 3/40 = 0.075

moles of HCl = 1 x 75 /1000 = 0.075

HCl + NaOH --------------------> NaCl + H2O

for 0.075 mol -----------------------> 2771.9 J

     1 mol reaction --------------> 2771.9 / 0.075 = 36958.67 J = 37 kJ

DHrxn = 37 kJ /mol

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.