Answer in General Chemistry for Dawson #169285
When 28.2 mL of 0.500 M H2SO4 is added to 28.2 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH 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.
“H_2SO_4 + 2KOH rightarrow K_2SO_4 + H_2O”
Concentration of “H_2SO_4” = 0.500M
Volume of “H_2SO_4” = 0.028L
No. of moles of “H_2SO_4” = “Concentrationtimes Volume”
= “0.028times 0.500”
= “0.014” mol
Concentration of KOH = 1M
Volume of KOH = 0.028L
Hence, number of moles of KOH = “0.028times 1”
= “0.028”
The mole ratio is “1:2”
So all the quantity is used both for “KOH” and “H_2SO_4”
“Mass = Densitytimes Volume”
“V_{total} = 56.4mL”
d = 1 g/mL
Hence, m = 56.4 g
“c= 4.2” J/gK
“Delta T” = 30.17 – 23.5
“Delta T = 6.67^ circ C”
“Q = ctimes Delta T times m”
= “4.2times 6.67times 56.4”
= “1579.98” J
“Delta H = dfrac{Q}{n}”
= “dfrac{1579.98}{0.04}”
= “39499.74” J
