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

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.

REACTION 1:

NaOH(s)=Na(aq)+OH(aq)

Data table1:table for the values of   the reaction   NaOH(s)=Na(aq)+OH(aq)

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 ...