Additionally the masses and volumes of solution were duplicated for all

Additionally the masses and volumes of solution were

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temperature rod had reached the same temperature as the solution before reaction. Additionally, the masses and volumes of solution were duplicated for all reactions.
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Data Tables: Table 1: Reaction 1 Reaction 2 Reaction 3 Mass of solid sodium hydroxide (grams) 1.960 2.0225 Volume of deionized water (mL) 100.00 50.00 Volume of 1.0 M HCl (mL) 49.5 49.5 Volume of 1.0 M NaOH (mL) 50.0 Initial temperature, Ti, °C 25.6 24.8 23.7 Final temperature, Tf, °C 30.1 36.0 30.5 Change in temperature ∆T, °C 4.5 11.2 6.8 Table 2: Reaction 4 Reaction 5 Reaction 6 Mass of solid sodium hydroxide (grams) 1.960 2.0264 Volume of deionized water (mL) 100.00 50.00 Volume of 1.0 M CH3COOH (mL) 50.00 50.0 Volume of 1.0 M NaOH (mL) 50.0 Initial temperature, Ti, °C 25.6 24.3 23.5 Final temperature, Tf, 30.1 35.2 30.0
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°C Change in temperature ∆T, °C 4.5 10.9 6.5 Calculations: Reaction 1: Mass of water =100.0g ΔT = 4.5 o C Specific heat = ( ¿ o Cg ) 4.18 J / ¿ ( ¿ o Cg )( 100.00 g )∗ 4.5 o C = 1880 J q water ( Joules )= 4.18 J / ¿ q water ( kJ )= 1.88 kJ Moles of NaOH = 0.049 mol q water ( kJ / mol )= 38.4 kJ / mol q reaction ( kJ / mol )=− 38.4 kJ / mol Reaction 2: Mass of water =99.5g ΔT = 11.2 C Specific heat = ( ¿ o Cg ) 4.18 J / ¿ ( ¿ o Cg )( 99.5 g )∗ 11.2 C = 4660 J q water ( Joules )= 4.18 J / ¿ q water ( kJ )= 4.660 kJ Moles of NaOH = 0.05056 mol Moles HCl= 0.0500 mol Limiting reactant: HCl q water ( kJ / mol )= 93.2 kJ / mol q reaction ( kJ / mol )=− 93.2 kJ / mol Reaction 3: Mass of water=99.5g ΔT = 6.8 C Specific heat= ( ¿ o Cg ) 4.18 J / ¿ ( ¿ o Cg )( 99.5 g )∗ 6.8 o C = 2830 J q water ( Joules )= 4.18 J / ¿ q water ( kJ )= 2.830 kJ
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Moles NaOH= 0.0495 mol Moles HCl= 0.0500 mol Limiting reactant=NaOH q water ( kJ / mol )= 57.2 kJ / mol q reaction ( kJ / mol )=− 57.2 kJ / mol Evaluating the Validity of Hess’s Law: Measured Δ H 3 =− 57.2 kJ Value of ∆H for Reac tion 3 bas ed on Hess’s law = ∆H2 – ∆H1 = 93.2 kJ / mol −(− 38.4 kJ / mol )=− 54.8
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