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Our aim in this experiment to measure the heats of reaction for three related exothermic reactions and to verify Hesss Law of Heat Summation

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Introduction

HEATS OF REACTION - HESS'S LAW INTRODUCTION: Our aim in this experiment to measure the heats of reaction for three related exothermic reactions and to verify Hess's Law of Heat Summation.By calculating temperature differences and calculating the energy lost or gained we are going to calculate the heats of reactions.Heats of reactions will be found by dividing the heat released to the moles of matter NaOH used. NaOH(s) --> Na+(aq) + OH-(aq) Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq) NaOH(s) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq) Hess's law states that energy changes are state functions. The amount of energy depends only on the states of the reactants and the state of the products, but not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction. http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html REACTION 1: NaOH(s)=Na(aq)+OH(aq) ...read more.

Middle

of 0,50M H + Cl trial2 22 26 26 26 25,5 25,5 50ml(�0,1) of 0,50M Na + OH, 50ml(�0,1) of 0,50M H + Cl trial3 22 25 25,3 25 25 25 50ml(�0,1) of 0,50M Na + OH, 50ml(�0,1) of 0,50M H + Cl Graph4:temperature - time graph of Na(aq) + OH(aq) + H(aq) + Cl(aq) = Na(aq) + Cl(aq) + H2O(l) for trial 1. Graph5:temperature - time graph of Na(aq) + OH(aq) + H(aq) + Cl(aq) = Na(aq) + Cl(aq) + H2O(l) for trial 5. Graph6:temperature - time graph of Na(aq) + OH(aq) + H(aq) + Cl(aq) = Na(aq) + Cl(aq) + H2O(l) for trial 6. CALCULATIONS FOR REACTION 2 Volume of 0,50M HCl used: 50ml(�0,1) Volume of 0,50M NaOH used: 50ml(�0,1) Mass of solution : 50g(�0,1)HCL+50g(�0,1)NaCl=100g(�0,1) Initial temperature of HCL solution:22,5C(r)(�0,1) Initial temperature of NaCl solution:23C(r)(�0,1) Final temperature of solution:24,2C(r)(�0,1) ?t=24,2C(r)(�?0,04) - 22,8 C(r)(�?0,04)=1,4(�?0,1) Final temperature of solution:25C(r)(�0,1) ?t=25C(r)(�?0,04) - 22,8 C(r)(�?0,04)=2,2(�?0,1) Final temperature of solution:25,5C(r)(�0,1) ?t=25,5C(r)(�?0,04) - 22,8 C(r)(�?0,04)=2,7(�?0,1) ?T average=2,1(�?0,1) Q=MC?T =100 g(�%0,1)*4,18*2,1(�%0,1)=877,8(�%0,2)j =0,9(�%0,2)kj 1000ml=1l =50 ml(�?0,2) =0,05l(�?0,2) =0,05l(HCL)*0,50mol/l(HCL) =0,025 mol HCL =0,05(�?0,2)l(NaOH)*0,50mol/l(NaOH) =0,025 mol(NaOH) (�?0,2) Heat of reaction=0,9(�%0,2)KJ/0.03(�?0,2)= - 30(�?0,4) REACTION 3 NaOH(s) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq) ...read more.

Conclusion

� Na+ + OH- Reaction 3 NaOH (s) + H+ + Cl- � Na+ + Cl- + H2O Volume H2O or HCl used (mL) 50 50 Mass of solid + container (g) 42 42 Mass of empty container (g) 41,4 41,4 Mass of solid used (g) 0,6 0,6 Initial water/HCl temperature (�C) 23 22,5 Final water/HCl temperature (�C) 25,6 27,6 Change in temperature (�C) 2,6 5,1 Reaction 2 Na+ + OH- + H+ + Cl- � Na+ + Cl- + H2O Volume NaOH solution (mL) 50 Volume HCl solution (mL) 50 Total solution volume (mL) 100 Initial temperature HCl solution (�C) 22,5 Initial temperature NaOH solution (�C) 23 Average temperature (�C) 22,8 Final temperature of mixture (�C) 24,9 Change in temperature (�C) 2,1 CONCLUSION: NaOH(s) --> Na+(aq) + OH-(aq) Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq) +____________________________________________ NaOH(s) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq) ?H1+?H2=-30+(-30) = -60= ?H3 But the ?H3 found in the end of the experiment is -50. Percent difference = ?H3-( ?H1+ ?H2)/ ?H3 *100 -50(-30+ -30)/-50*100 =? 20 In the end the heats of reactions are measured and the Hess law is verified. CHEMISTRY EXPERIMENT HEATS OF REACTION - HESS'S LAW Beril Kologlu 41831 11-F ...read more.

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