How do I calculate the partial pressure of HCl?

Develop a procedure for determining the molar enthalpy of neutralization for the reaction of NaOH and HCl?

• No clue how to do this. The lab materials include a constant-pressure calorimeter and a Styrofoam cup. A temperature probe will be used for measuring change in temperature. q(system) = -q(surroundings) The small heat capacity of the cup will be neglected in the experiment. You will be provided with a total of 150 mL of 2M NaOH and 150 mL of 2M HCl. Perform 3 determinations and average your final results, Calculate percent error.

• Answer:

  measure precisely as you can 50 ml each of NaOH & HCl use the probe to measure the temperature before & after mixing when you mix them stir well weigh the amount of salt water produced calculate the heat released by the reaction , using the formula : dH = m C dT dH = (mass of salt water) )pure water's specific heat of 4.184J/g-C)(dT) =========================== your reaction was on a small 50 ml scale: 0.050litres @ 2 mol/litre = 0.10 mole experiment you used 0.1 mol HCl & 0.1 mol NaOH --> produced 0.1 mol H2O for determining the molar enthalpy of neutralization: divide (your joules released to the water) by (0.1 mole ) = joule/mole yoiu should have enough for 3 trials ====================================== % error is determined upon how close you can get to the standard dH : HCl(aq) & NaOH (aq) --> H2O(l) & NaCl (aq) dH rxn = dHf prod - reactants dH = [Hf H2O & Hf NaCl(aq)] - [Hf HCl(aq) & Hf NaOH(aq)] dH = [-285.8 & -407.1] - [-167.2 & -469.6] dH = -692.9 - (-636.8) dH = -56.1 kJ again % error would be how close did you get to -56.1kJ/mole neutralized