|Jun20-06, 08:33 PM||#1|
I solved for the heat of neutralization already, I thought, but now, I got the final temperature after reaction is 121.75 oC which I think is too big.
I start with initial pH = 2 and final the conc H+ = 10^-(pH=2) then I find the mole by multipling the conc with the tank volume (400m3), the I find the mass by multiply it with MW.
To neutralise 1 mole of H+, I need 1 mole of OH-, so I balance the mole of OH- needed with the mole of H+ I calculated justnow. The I got the mass and volume of NaOH needed to neutralize the pH 2.
The heat of neutralization is -55.9 kJ/mol. The energy balance is (m.Cp.del_T)in = (m.del_H) + (m.Cp.del_T)out
But since i use the initial temperature and reference temperature both also 25 oC, so the (m.Cp.del_T)in is cancelled. What left is only
-(m.del_H) = (m.Cp.del_T)out
The part that confused me here is what m should I use for these m.del_H and m.Cp.del_T. I obtained the answer stated ealier (121.75 oC) by using the mass for m.del_H is the mass of water produced and the mass for m.Cp.del_H is the mass of the total mass of NaOH+H2SO4 that reacted. Is this correct? If yes, is the temperature I got reasonable? If not, whats my mistake?
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