I am pretty sure I did this completely wrong. When 25.8 mL of 0.500 M H2SO4 is added to 25.8 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH...

Question

of this reaction. (Assume that the total volume 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·°C.) Do not answer in scientific notation.

I got -0.0558 kJ/mol H₂O

Answer 1

Hello!

Total volume of your solution = 25.8 mL + 25.8 mL = 51.6 mL

Mass of solution (M) = volume • density

51.6 mL • 1.00 g/ml = 51.6 g

Change in temperature (ΔT) = 30.17 º C - 23.5 º C = 6.67º C

Specific heat capacity(c) = 4.184 j/g•ºC

ΔH = -q

= -mcΔT

= - (51.6 g) • (4.184 J/g•ºC) • (6.67ºC)

= -1440.02 J

= -1.44 kJ

The fact that the answer is negative shows that the reaction is exothermic.

Hope this helps!