Heidi Duncan 11/24/13 Exothermic and Endothermic Reactions Lab The purpose of this lab is to observe how heat is released or absorbed with different chemicals. Data Table 1 – HCI and NaOH Trial 1 Trial 2 Avg volume 1. 0 M HCl(ml) 25 Volumel . 0 M NaOH (ml) Ti of HCI before mixing 20 Ti of NaOH before mixing( Average Ti before mixing( Tf of mixture ) 26 T) 6 Specific Heat O/g) 4.

184 Heat, q 0) 1255. 2 Data Table 2- NH4 N03 and H20 Mass of NH4 N03 (g) 12 11. 93 Volume of H20 ( ml) Ti of H20 ) -3 -2 Specific Heat0/g) -2405.

8 -2301 . 2 Heat per gram O/g) -200 -193 -197 Data Table 3- NaOH and H20 Mass of NaOH (g) 7. 99 8. 6 Volume of H20 (ml) 50 Ti of H20 21 Tf of mixture 55 29 32 6066. 8 759 -831 795 Reaction 1 – HCI and NaOH 1 .

Determine the average initial temperature for each trial by averaging the initial temperature of the NCI and the NaOH solution before mixing. Record as Average Ti before mixing in Data Table 1. Substance Trial 1 (Ti) Trial 2 (Ti) Average (T’) HCI NaOH 20X 20 = 20 2.

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Calculate the change in temperature T by substracting the average initial temperature from the final temperature of the mixture (T = Tf -Ti) Record in Data Table 1. Substance Tf HC’ & NaOH rntx 26-20=6 3. Did the temperature rise or fall as the 2 solutions were mixed?Explain this in term of heat transfer. -The temperature rose as the 2 solutions were mixed because they neutralized. 4.

Calculate the heat of neutralization (q in Joules) for the reaction using equation (q =m X T X s). Assume that the density and specific heat of the solution are the same as that of pure water (density of water = 1. 0 g/ml). Record in data table 1. Calculate the average heat (q) by averaging trials 1 and 2. Record in data table 1.

Q=? S ??”4. 184 J/g *Both Trials had the same results 5. Classify the reaction as either exothermic or endothermic. Give evidence for your answer. The reaction is exothermic because the temperature rose. Reaction 2 – NH4 N03 and H20 1. Calculate the change in temperature (T) for each trial.

Record in Data Table 2. Trial -3 -20= -23 -2- 20 -22 this in terms of heat transfer. -The temperature of the water fell because the NH4N03 absorbed the heat of the water. 3. Calculate the heat (q in Joules ) for the reaction. Record q for each trial in data table 2.

Trial 1 M=25g S=4. 184J/g -3 – 20 ??”23 -r=2 -20=-22 T??”22. 5 -2405. 8] -2301. 2J -2353. 5 4.

Calculate the heat absorbed or released per gram of solute added to the water (in oules/g) record in data table 2.Calculate the average heat per gram by averaging trials 1 and 2. Trial 1 Trail 2 -2405. 8J/12g -2301. 2J/11. 93g -2353. 5 J 1. 965 g -200 vg -193 vg -197 vg 5.

Classify the reaction as either exothermic or endothermic. Give evidence for your The reaction was endothermic because the temperature fell. 6. What answer. – could be some possible uses for a chemical such as ammonium nitrate? -A possible use for ammonium nitrate would be in ice packs.

Reaction 3- NaOH and H20 1 . Calculcate the change in temperature T for each trial. Record in data table 3. Trial 50=29 55-23 2.Did the temperature of the water rise or fall when the NaOH was added. -The temperature of the water rose because the NaOH added heat to the water. 3. Calculate the heat (q in Joules) for the reaction.

Record for q in data table 2. Trial 1 avg M=50g -r=50- 21 -r=30. 5 6066. 8 J 6694. 4 J 6380. 6 J 4. Calculate the heat absorbed or released per gram o fsolute added to the water ( in joules ) Record in data table 2.

Calculate the average heat per gram by averaging 6066. 8 J 7. 99 g 6694. 4 J / 8. 06 g 6380.

6 J / 8. 025 g 759 vg 831 vg 795 vg answer. -The reaction was exothermic because the temperature rose.