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

Answers

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 moles

dh=-(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