PART II. HEAT OF NEUTRALIZATION OF STRONG ACID, HCL AND STRONG BASE, NAOH Balanc
ID: 558702 • Letter: P
Question
PART II. HEAT OF NEUTRALIZATION OF STRONG ACID, HCL AND STRONG BASE, NAOH Balanced reaction: 1. Temperature of 50 mL of 2.0 M HCI before mixing: 2. Temperature of 50 mL of 2.0 M NaOH before mixing: 3. Temperature of 100 mL of solution after mixing T- 23· T-31.5c 4. Assume that the density (1.00 g/ml) and specific heat of the solution (4.184 J/g"C) are the same as pure water. The heat gained by the solution is uwohution (mass, g) (4.184 J/g C) (T T) 2994 10o Using the heat capacity of the calorimeter in part I, the heat gained by the calorimeter is 5.lealionnete(heat capacity) (T-T) The total heat of neutralization, - Calculation 7. Moles of acid reacted 0 mole Calculation mole Moles of water produced Calculation 8. 11/mole 9. Calculate AHofneutralization per one mole ofwater formedExplanation / Answer
moles of acid = molarity* volume iin L= 2*50/1000 = 0.1 = moles of NaOH
the reaction between HCl and NaOH is HCl+ NaOH------>H2O+NaCl
1 mole of HCl reacts with 1 mole of base to give 1 mole of water
0.1 mole of HCl reacts with 0.1 mole of NaOH to give 0.1 mole of water
heat gained by solution = mass of solution* specific heat of water* change in temperature= 100*4.184*(37.5-23.4) =5899.44 joules
heat gained by calorimeter= heat capacity of calorimeter* change in temperature = 4.18*(37.5-23.4) =58.94 joules
total heat of neutralization = -(heat gained by solution+ heat gained by calorimeter)=-(5899.44+58.94) joules-5958.4 joules
heat of neutralization/mole of water = -5958.4/0.1= -59584 J/mole=-59.584 Kj/mole
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