Graph1:the graph of temperature- time graph of NaOH(s)=Na(aq)+OH(aq) for trial1.
Graph2:the graph of temperature- time graph of NaOH(s)=Na(aq)+OH(aq) for trial2.
Graph3:the graph of temperature- time graph of NaOH(s)=Na(aq)+OH(aq) for trial3.
CALCULATIONS FOR REACTION 1
H2O used: 50ml(±0,1)
mass of average NaOH used: 0,6(±0,01)
Mass of solution : 50,0g(±℅0,2)HCL+0,6(±℅1,6)NaOH=50,6g(±℅1,8)
Initial temperature of HCL solution:23C®(±0,1)
Final temperature of solution:26®(±0,1) Δt=26®(±℅0,04) – 23 C®(±℅0,04)=3(±℅0,1)
Final temperature of solution:27,5C®(±0,1) Δt=27,5C®(±℅0,04) – 23 C®(±℅0,04)=4,5(±℅0,1)
Final temperature of solution:23,5®(±0,1) Δt=23,5®(±℅0,04) – 23 C®(±℅0,04)=0,5(±℅0,1)
ΔT average=2,6(±℅0,1)
Q=MCΔT
=50,6 g(±%0,03)*4,18*2,6(±%0,1)=549,9(±%0,2)j =0,6(±%0,2)kj
0,6(±℅1,6)g NaOH /40g =0,02mol(±℅1,6)g
Heat released= 0,6(±%0,2)kj/0,02mol(±℅1,6)g =30 kj/mol
REACTION 2
Na(aq) + OH(aq) + H(aq) + Cl(aq) = Na(aq) + Cl(aq) + H2O(l)
Data table2:table for the values of the reaction Na(aq) + OH(aq) + H(aq) + Cl(aq) = Na(aq) + Cl(aq) + H2O(l)
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®(±0,1)
Initial temperature of NaCl solution:23C®(±0,1)
Final temperature of solution:24,2C®(±0,1) Δt=24,2C®(±℅0,04) – 22,8 C®(±℅0,04)=1,4(±℅0,1)
Final temperature of solution:25C®(±0,1) Δt=25C®(±℅0,04) – 22,8 C®(±℅0,04)=2,2(±℅0,1)
Final temperature of solution:25,5C®(±0,1) Δt=25,5C®(±℅0,04) – 22,8 C®(±℅0,04)=2,7(±℅0,1)
ΔT average=2,1(±℅0,1)
Q=MCΔT
=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)
Data table3:table for the values of the reaction NaOH(s) + H+(aq) + Cl-(aq) → H2O + Na+(aq) + Cl-(aq) .
Graph7:temperature – time graph of NaOH(s) + H+(aq) + Cl-(aq) → H2O + Na+(aq) + Cl-(aq) for trial 7.
Graph8:temperature – time graph of NaOH(s) + H+(aq) + Cl-(aq) → H2O + Na+(aq) + Cl-(aq) for trial 8.
Graph9:temperature – time graph of NaOH(s) + H+(aq) + Cl-(aq) → H2O + Na+(aq) + Cl-(aq) for trial 9.
CALCULATIONS FOR REACTION 3
Trial 1
Volume of 0,25M HCl used: 50ml(±0,1)
mass of average NaOH used: 0,6(±0,01)
Mass of solution : 50,0g(±℅0,2)HCL+0,6(±℅1,6)NaOH=50,6g(±℅1,8)
Initial temperature of HCL solution:22,5C®(±0,1)
Final temperature of solution:28,2C®(±0,1) Δt=28,2C®(±℅0,04) – 22,8 C®(±℅0,04)=5,4(±℅0,1)
Final temperature of solution:28,9C®(±0,1) Δt=28,9C®(±℅0,04) – 22,8 C®(±℅0,04)=6,1(±℅0,1)
Final temperature of solution:25,7C®(±0,1) Δt=25,7C®(±℅0,04) – 22,8 C®(±℅0,04)=2,9(±℅0,1)
ΔT average=4,8,(±℅0,1)
Q=MCΔT
=50,6 g(±%0,03)*4,18*4,8(±%0,1)=1015,2(±%0,2)j =1(±%0,2)kj
0,6(±℅1,6)g NaOH /40g =0,02mol(±℅1,6)g
Heat released= 1(±%0,2)kj/0,02mol(±℅1,6)g =50 kj/mol
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)
ΔH1+ΔH2=-30+(-30) = -60= ΔH3
But the ΔH3 found in the end of the experiment is -50.
Percent difference = ΔH3-( ΔH1+ ΔH2)/ ΔH3 *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 Koloğlu
41831 11-F