500.0 mL of 0.220 mol/L HCl(aq) was added to a high quality insulated calorimeter containing 500.0mL of 0.200 mol/L NaOH(aq).Both solutions had a density of 1g/mL & a specific heat of 4.184 J/g.K.The calorimeter had a heat capacity of 850 J/C.Temperature of entire system rose from 25.60 C to 26.70 C.Calculate dH (kJ/mole) of NaOH(aq).
Here is my approach for:
As you said earlier, the q reaction is:
q = mcdT + CdT
= (500g+500g)(4.184)(1.1) +850.0(1.1)
= 5537.4 J
Since neutralization reaction occurs,heat evolves,therefore
q reaction = -5537.4 J
Calculating the number of moles of NaOH:
nNaOH = 0.200mol/L X 0.5L = 0.1 mol
q reaction = nNaOH X dH
dH = (-5537.4J)/0.1mol = -55374 J/mol
= -55.4 kJ/mol,since answer asks for 1 decimal place.
Is my steps and solution correct? Can Dr.Bob clarify? Thank you
Chemistry - DrBob222, Monday, March 26, 2012 at 12:51pm
That looks ok to me.