# Temperature vs time for weak acid and naoh

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Temperature vs Time for Weak Acid and NaOH temperature of mixture (°C) Linear (temperature of mixture (°C)) Linear (temperature of mixture (°C)) Time (s) Temperature (°C) Calculations: q = VdCp∆T heat lost by hot water = 50mL x 1.00gmL -1 x 4.18Jg -1 K -1 x (31.25-41.9)K = -2225.85 J heat gained by cold water = 50mL x 1.00gmL -1 x 4.18Jg -1 K -1 x (31.25-21.9)K = 1954.15J |total heat lost| = |total heat gained| 2225.85 J = 1954.15 J + Cp(31.25-21.9) 2225.85 J – 1954.15 J = 9.35Cp 271.7 J/9.35 = Cp = 29.1 JK -1 3
∆T of acid = 26.8°C–24.5°C= -2.3°C ∆T of base = 26.8°C–22.3°C = 4.5°C q = V c dC p ∆T’+C p cal ∆T’ 1. q p = ∆E + P∆V q p of acid = C p ∆T = (4.18Jg -1 K -1 )(-2.3)K = -9.6 J q p of base = C p ∆T = (4.18Jg -1 K -1 )(4.5)K = 18.8 J n HCl = (1.000mol/L)(0.05L) = 0.05mol n NaOH = (0.9000mol/L)(0.05L) = 0.045mol 2. q = |V acid dC p ∆T| + |V base dC p ∆T| + |C p ∆T| = |(50mL)(1.00gmL -1 )(4.18Jg -1 K -1 )(-2.3)K| + |(50mL)(1.00gmL -1 )(4.18Jg -1 K -1 )(4.5)K| + |29.1JK -1 )(9.35)K| = 1693.23J ∆H 1 = -q/n = -1693.23J/0.045mol = -37.6kJ 3. HA + OH - +H 2 O A - + 2H 2 O K A = 1.35x10 -5 0.997 0 0 0 0 [HA] = 0.997-x [H 3 O + ] = x [A - ] = x 1.35x10 -5 = ( x )( x ) 0.997 x = x 2 0.997 x √x2 = √1.35x10 -5 = 3.7x10 -3 mol/Lx0.05L = 1.85x10 -4 = [H 3 O + ] Heat released by reaction a. = -37kJ/mol x 1.85x10 -4 mol/L = -6.8x10 -3 kJ ∆H 2 = -6.8x10 -3 kJ 4. 2H 2 O H 3 O+ + OH - ∆H 1 = 37.6 kJ (flipped) HA + OH - + H 2 O A - + 2H 2 O ∆H 2 = -6.8x10 -3 kJ HA + H 2 O H 3 O + + A - ∆H 3 = 37.6 kJ (∆H 1 + ∆H 2 ) 5. a) ∆G° = -RTlnKA = -(8.314x10 -3 )(298K)(ln 1.35x10 -5 ) = 27.8 kJ b) ∆G° = ∆H 3 - T∆S° ∆G° - ∆H 3 /-T = ∆S° = 27.8kJ – 37.6kJ/-21.6 = 0.454kJmol -1 K -1 Discussion: This experiment allowed the enthalpies for HCl, HA and the ∆H of ionization for the reactions, ∆G° and ∆S°. The HA used in the calculations was monochloro acetic acid. By extrapolating the graph back to time zero, the highest temperature reached from the time the liquid (either hot water, HCl or HA) is added to the calorimeter. This temperature was used as T 2 in calculating the Cp of the calorimeter as 29.1J. This value is
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