Homework #37

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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 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

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