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/newconneet.mheducation.com/flow/connect.html AMyCU Login udy sign In "Sign in to Office 365 m Blackboard Learn Capsiml Business ment 10 Chapter 6 Sa Check my work Enter your answer in the provided box. When 29.4 mL of 0.500 M H2SO4 is added to 29.4 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 3 is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water) (d for water = 1.00 g/mL; c for water = 4.184 J/g.oC) 0.170. Calculate AH of this reaction. (Assume that the total volume kJ/mol H2oExplanation / Answer
moles of acid H2SO4 = 29.4 x 0.50 / 1000 = 0.0147
moles of NaOH = 29.4 x 1.0 / 1000 = 0.0294
H2SO4 + 2 NaOH ------------------------> Na2SO4 + 2H2O
1 2
0.0147 0.0294
here both are consumed
moles of water fomred = 0.0294
total volume = 29.4 + 29.4 = 58.8 mL
mass of solution = volume x density
= 58.8 x 1
= 58.8 g
dT = 30.17 -23.50 = 6.67 oC
Q = 58.8 x 4.184 x 6.67
Q = 1641 J
molar enthalpy of neutralization = - Q / n
= -1.641 / 0.0294
= - 55.8 kJ / mol
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