that took forever to figure out
change of heat of reaction= Vsolution/moles of limiting reactant X specific heat X change in temp X 1kj/1000j
When 25.0 mL of 1.0 M H2SO4
is added to 50.0 mL of 1.0 M NaOH at 25.0 °C in a calorimeter, the temperature of the aqueous solution
increases to 33.9 °C. Assuming the specific heat of the solution is 4.18 J/(g·°C), that its density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate H (in kJ) for the reaction:
H2SO4(aq) + 2 NaOH (aq) 2 H2O (l) + Na2SO4
(aq)
3 answers
25ml+50ml=75ml *(1g/ml)= 75g
q= m*c*delta T
q= (75g) (4.18) (+8.9)
q= 2790.15 J = delta H
q= 2.79 KJ
25ml H2SO4
( x / 0.25ml) = 1 M H2SO4
x= 0.25 mol H2SO4
2.79KJ/ 0.25 mol H2SO4 = 11.16 KJ/mol H2SO4
delta H= - 11.2 kJ/mol
I'm not sure why it's negative but I hope this helps.
q= m*c*delta T
q= (75g) (4.18) (+8.9)
q= 2790.15 J = delta H
q= 2.79 KJ
25ml H2SO4
( x / 0.25ml) = 1 M H2SO4
x= 0.25 mol H2SO4
2.79KJ/ 0.25 mol H2SO4 = 11.16 KJ/mol H2SO4
delta H= - 11.2 kJ/mol
I'm not sure why it's negative but I hope this helps.
It is negative because the temperature increases, therefore heat is liberated. This means it would be exothermic (delta H is -)
2 answers