Expt: H Enthaply

Question 1

Endothermic

Answer 1

A chemical reaction that absorbs heat from the surroundings is called endothermic, and as a result, one feels that the surroundings become cooler.

If the reaction occurs at a constant pressure, the heat flow caused by the reaction, is numerically equal to the change in the systems enthalpy over the course of the reaction, if the reaction is exothermic the systems enthalpy increases, and 🔺Hsys > 0.

Question 2

Exothermic

Answer 2

Reaction that generates heat is called exothermic; out of the system (🔺Hsys is negative) into the surroundings (🔺Hsurr is positive), and the surroundings become warmer

If the reaction occurs at a constant pressure, the heat flow caused by the reaction, is numerically equal to the change in the systems enthalpy over the course of the reaction, that is,qrxn =-🔺Hsys. Hence, if the reaction is exothermic, the systems enthalpy decreases, and 🔺Hsys < 0.

Question 3

Constant pressure calorimeter

Answer 3

Is it device used to measure the temperature change of a system over the course of a reaction, a phase change or the dissolution of solute in water, which are processes that can either absorb or release heat.

Question 4

What experimental condition must hold to determine the values of 🔺H from calculated heat Transactions occurring inside a thermos calorimeter?

Answer 4

Experiment performed at a constant internal pressure

Question 5

How is the heat absorbed by the solution (qsoln) inside the calorimeter related to the heat released by the reaction (qrxn)?

Answer 5

qrxn = -qsoln

Question 6

The dissolution of NaNO2 (s) in water is endothermic. What happens to the temperature of the water as NaNO2 dissolves?

Answer 6

The temperature of the water decreases as NaNO2 dissolves 

Question 7

Use your lowest value measured for your final temperature and calculate 🔺t. Calculate the heat of the reaction, qrxn, in joules, using the correct sign to indicate if the heat is being absorbed or given off. Assume that the solution density is 1.02 g mL^-1. 

Answer 7

q_soln = m x C x 🔺t

m = 100 mL x 1.02 g/mL = 102 g

= 102g x 4.19 °C J x -120 °C = -512.856 J

q_rxn = -q_soln
= -(-512.856) = 512.856 = 513 J

Question 8

Calculate 🔺H^0soln (Na2SO4 • 10 H2O)

Answer 8

n = m/M —> 5.34g/322.187338 = 0.0165742081 mol

🔺H_soln —> 0.513KJ/0.0165 mol = 30.94301682 = 30.9 KJ/mol

Question 9

Calculate your experimental value for 🔺H^0hydration (Na2SO4) applying hess’s law. Show the addition of the appropriate net ionic equation is used to obtain 🔺H^0hydration 

Answer 9

Na₂SO₄ —> 2 Na⁺ + SO₄^2- 🔺H_soln = -4.43 KJ/mol

2 Na⁺ + SO₄^2- + 10 H_2- —> Na₂SO₄ • 10 H_2- + 30.9 KJ/mol

🔺H_hydration = -4.43 - (-30.9) = 26.47 KJ

Question 10

Complete the table below with the values of the standard formation of enthalpies of Na2SO4(s). And H2O(l).

Answer 10

Compound.  🔺Ht^0 (kJ mol^-1)
Na2SO4 -1387
H2O -285.8
Na2SO4 • 10 H2O (s) -4327.3

Na₂SO₄ —> Na₂SO₄ • 10 H_2-

🔺H_Na2SO4 = 🔺H_f products - 🔺H_f reactants

= -4327.3 - (-1387 + (10 • -285.8))
= -4327.3 - (-4245)
= -82.3 KJ/mol

Question 11

Calculate the percent error of your experimental value for 🔺H^0hydrarion (Na2SO4) with respect to the reference value of 🔺H^0hydration (Na2SO4) calculated from the equation.

Answer 11

% Error = experimental value - reference value / reference value x100

= 26.47 - (-82.3)/-82.3 x100
=-132.162819
= -132%

Question 12

Results (NaOH + HCl)
Calculate the heat of the reaction, qrxn, in Joules. Use the correct sign to indicate if the heat is being absorbed or given off. Assume that the solution density is 1.00 g mL^-1

Answer 12

q_rxn = m x C x 🔺t

100 g x 4.19 °C J x 6.00 °C = 2514 J

2514J/1000KJ = 2.514 KJ

q_rxn = -q_soln
= -(2514) = -2514 J

Question 13

Write the balanced molecular equation for the neutralization reaction of NaOH and HCl. Calculate the 🔺H^0 neutralization (in kJ mol^-1) per mile of water formed. 

Answer 13

HCl + NaOH —> NaCl + H2O

n = c • v
= (1.00mol/L)(0.0500L)
=0.0500 mol
Therefore, 0.0500 mol of H2O formed.

n = c • v
= (1.25mol/L)(0.0500L)
=0.0625 mol

🔺H_{neutralization} = -2.514 KJ / 0.0500 mol = 50.28 KJ/mol

Question 14

Results (NaOH + CH3COOH)
Calculate the heat of the reaction, qrxn, in joules. Use the correct signed indicate if the heat is being absorbed or given off.

Answer 14

100mL x 1.00g/mL = 100g

qrxn = 100 x 4.19 J °C x 5.1°C = 21369 J

q_rxn = -q_soln
= -(2.1369) = -2.1369 KJ

CH3COOH + NaOH —> NAOOCH3 + H2O

n = c • v
= (1.00mol/L)(0.0500L)
=0.0500 mol
Therefore, 0.0500 mol of H2O formed.

n = c • v
= (1.25mol/L)(0.0500L)
=0.0625 mol

🔺H_{neutralization} = -2.1368 KJ / 0.0500 mol = 42.73 KJ/mol

Question 15

Your results should have revealed that the neutralization of HCl with NaOH is more significantly exothermic then the neutralization of CH3COOH with NaOH.
Explain this experimental fact based on the type a number of bonds that are formed and broken in each neutralization reaction.

Answer 15

With NaOH & HCl. HCl is a strong acid and therefore they will fully dissolve = it’s more exothermic and the bonds will completely break before OH^- is added to the system.

With acetic acid it’s a weak acid, meaning Some energy is absorbed by the system to break the existing H-A bonds.

Question 16

A student performed a calorimeter calibration using a thermos tell him calorimeter Of the type used in this experiment. First, the student added 50.0 mL of water at 8.4°C to the calorimeter, Which also was at the same temperature. Then, the student added 40.5 mL of water at 81.5°C. After mixing, the final temperature was found to be 40°C. Determined he gained by the calorimetre (qcal) in the calibration run

Answer 16

Hot water
m = d • v = 1.00g/mol • 40.5mL = 40.5g
🔺t = tf - ti
= 40.5-82.5 =-41.5°C

qhot = 40.5g x 4.19°C J x -41.5°C
= -7042.3425 J
=-7.04x10^3 J

Cold water
m = d • v = 1.00g/mol • 50.4mL = 50.4g
🔺t = tf - ti
= 40.0-8.4 =31.6°C

qhot = 50.4g x 4.19°C J x 31.6°C
= -6673.166 J
=6.67x10^3 J

qcal = qcold + qhot
=7.04x10^3 -6.67x10^3
=369.1809 J
qcal = 3.7x10^2 J

Question 17

Calculate the heat capacity of the calorimeter (Ccal) Do used in the mock calibration described in the previous question. 

Answer 17

qcal = Cp • 🔺t
Cp = qcal/🔺t

🔺t = 40.0-8.4 = 31.6°C
3.7x10^3 J / 31.6°C = 11.708… = 12 J/°C

Question 18

How much heat is required to raise the temperature of 100 g Of water by 10°C? How much heat is required to raise the temperature of a calorimeter by 10°C?
Based on these values, what is Valid for you to ignore the heat absorbed by the calorimeter in this experiment?

Answer 18

q = m x Cp x 🔺t
=100g • 4.19J°C • 10°C
=4190 J

q = Cp • 🔺t
= 12J/°C • 10°C
= 120 J

120 J/ 4190 J x 100 = 2.863… =2.9%

Yes b/c the heat required to raise the temp of water by 10°C compared to the heat absorbed by the calorimeter for the same 10°C, is a very small number.