heat of solution: I will have this experiment on next lab and want to be prepare
ID: 837943 • Letter: H
Question
heat of solution:
I will have this experiment on next lab and want to be prepared for it but don't know how to calculate
some things like molarity, initial and final temperature, change in temperature, heat of neutralization per mole and so on :(( .. I will be grateful for explaining to me how to solve each problem step by step, so will be able to do it on my own later. Thank you!!!!
page 60: and without piecing the bottom, record the temperature of the HCL solution. Rinse the termometer and then record the temperature of the NaOH solution in the Erlenmeyer glask. The two temperatures must be within 2C before starting the experiment. Record the temperatures and take the acerage as the initial temperature. Using a clean 50ml graduated cylinder, pour 25ml of NaOH solution into the Styrofoam cup calorimeter... so on so on.....Record the the maximum temperature that was observed as final temperature. Determine the change in temperature, (triangle)T. Assuming that the density of the solution was 1,00 g/ml and the final colume of the solution was 50,0ml, determine the mass of the solution. Given that the specific heat of water is 4.18 J/g C, calculate the (triangle)H for theneutralization reaction. Finally calculate the (triangle)T for the neutralization reaction per mole for H+ and OH- ions reacting.
Repeat the procedure for trail 2 and after trial 2 calculate the average (triangle)H for the neutralization reaction per mole of H+ and OH- ions reacting.
The heat of neutralization will be determined in the reaction of HCl and NaOH. Obtain 54.0 mL of water in a 100 mL graduated cylinder and pour it into a 125 mL Erlenmeyer flask. Add 6.0 mL of 6.0 M HCl to the flask using a 10 mL graduated cylinder and swirl it until it is thoroughly mixed. Make sure to label the flask. Add 54 mL of water to a second labeled 125 mL Erlenmeyer flask. To this flask add 6.0 mL of 6.0 M NaOH using a 10 mL graduated cylinder. Again swirl the flask until the solution is thoroughly mixed. Calculate the molarity of the new HCl and NaOH solutions. Assemble a simple Styrofoam cup calorimeter using 2 Styrofoam cups. Place the nested cups into a 400 mL beaker for support. Add 25 mL of the freshly prepared HCl solution to the Styrofoam cup calorimeter. Carefully place the thermometer in the Styrofoam cup calorimeterExplanation / Answer
Molarity of NaOH solution = (Initial volume of NaOH * Initial molarity of NaOH )/ (Final volume) = (6* 6) / (6 + 54) = 0.6 M
Molarity of HCl solution = (Initial volume of HCl * Initial molarity of HCl )/ (Final volume) = (6* 6) / (6 + 54) = 0.6 M
Let say the initial temperature of HCl = T1 = 28 C
Let say the initial temperature of NaOH = T2 = 30 C
SO average initial temperature = (28 + 30) /2 = 29 C
Let say final temperature = T = 50 C
Change in temperature = (triangle) T =final temperature -average initial temperature = 50 - 29 = 21 C
Mass of solution = volume of solution * density of solution = 50 * 1 = 50 gm
Heat of neutralization = Mass of solution * ((triangle) T ) * specific heat = 50 * 21 * 4.18 = 4389 J
Number of mole sof H+ present in the 25 mL solution = volume (in litres)* Molarity of HCl solution = (25/1000)* 0.6 = 0.015 moles
Heat of neutralization per mole = Heat of neutralization/ Number of mole sof H+ present in the 25 mL solution = (4389/0.015 ) = 292600 J/mol = 292.6 KJ/mol
Average Heat of neutralization per mole = we take an average of two Heat of neutralization per mole for the 2 trials.
If the final temperature is greater than average initial temperature , then the neutralization reaction is exothermic.
If final temperature is lesser than average initial temperature endothermic
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