Homework #37greenspun.com : LUSENET : UR General Chemistry : One Thread |
I don't understand how to calculate the dH for this problem. Do I use dH = mCdT ? Is m = 100g ? Is C = 4.18 ? Is dT = 0.8 ? I have used this process for this problem and gotten the wrong answer once I tried to solve for dH in kJ/mol for AgCl. What step am I doing wrong?
-- Anonymous, February 07, 2000
You are doing everything correct, except you are neglecting to read the entire question. The answer has to be per moles of AgBr. Moles of AgBr produced = (.05 L)(.1M) = .005 molesdh=-(4.18)(100grams)(23.4C-22.6C)=334J -334J/.005moles = -66KJ
Hope that helps.
-- Anonymous, February 07, 2000
You did the first part right, dH=mCdT, and all your values are right. THis should give you 330J.The problem assumes that .05L * 0.1 mol/L= 0.005mol of AgNO3 and HCl are reacted. Therefore, 0.05mols of AgCl is produced becuase there is a 1:1 mol ratio between reactants.
dH = (330J)/(0.005mol)= 66000J = 66kJ **This value is then made negative because you have an increase in temp. and the reaction exothermic. dH = -66kJ
-- Anonymous, February 07, 2000