Question

Use the Henderson-Hasselbalch Equation to calculate the pH of a buffer solution that 10 0.050 Min benzoic acid (HC,1,0,) and 0.150 Min sodium benzoate ac or benzoic acid, K, - 6.5 * 10. (25 points) 2. Consider the combustion of propane gas: (25 points) CH, (g) + 5 0,(9) -- 3 CO2(g) + 4 H20 (g) ; AHU - -2044 kJ (a) Calculate the entropy change in the surroundings when this reaction occurs at 25 °C. (b) Determine the sign of the entropy change for the system.

Answer 1

Ang 0 Molavity of benzoic auid = 0.050M Molavity of Sodium benzoate= 0.150 M Ka, for benzoic acid = 6.5 xios pka = -log ka = -log (6.5X105) = -log 6.5+ leg 10s) - [0.8129 - 5] 4-1871 1 pka = 4.191 According to Henderson - Hasselbalch equation pH = pka + log [ Salt] [ Acid] pH = pka + log Sedium benzoate] į ben zait acid ] = 4.19 + 109 [ 015) [0.050) 4.19 + log 3 4.19 + 0.477 4.19 +0.48 IPH = 4.67 pH of the buffer solution is 4.67

Art @ Combustion of propane gas - C₃Hq (g) + 50 g) 3 co go + 4H00 (9) . Hoxn=-2044 KJ Entropy change in swerounding = ? Temperatwee = 25°c =(25+2+3.JK 298k As surrounding - Temp. =-(-2044 kJ) 298 = +6.8590K5 4 Sowetourding = +6.86 kJ The value of delta H is negative because the exothermic. The value of delta s is obtained a positive value. An exothermic reaction has always detta s a positive value i.e. heat is released to the surrounding,