Date: Section: Name: Prelaboratory Assignment: Investigation of Hess\'s Law Thro

Question

Date: Section: Name: Prelaboratory Assignment: Investigation of Hess's Law Through Calorimetry 1. What is the molarity of a solution prepared by diluting 25 ml of 6 MHCl with 75 ml of deioinized water? 2. a) Calculate the amount of heat required to raise the temperature of 87.30 g of water from 25°C to 100°C. (c H,0-4.18 J/g c) b) Calculate the amount of heat involved in the conversion of 87.30g of water at 100°C to steam at 100°C. (AH aperication 40.7 kJ/mol) c) Use Hess's law to calculate the amount of heat involved in converting 87.30 g of liquid water at 25°C to steam at 100°C

Explanation / Answer

1.Volume before dilution =25ml, volume of deionized water added= 75ml

Total volume after dilution = 25+75= 100ml

Moles of HCl before dilution = molarity* volume inL= 6*25/1000

This moles remain the same even during diltution. Concentration after dilution = 6*25/1000/(100/1000)= 1.5M

2. heat required to rise the temperature of water= mass of water* specific heat* temperature difference = 87.3*4.18*(100-25)= 27369 joules=27.369 KJ

3. water at constant temperature of 100deg.c will be converted from liquid to steam by supplying latent heat and heat required = moles of water* heat of vaporization =( mass of water/ molar mass of water)*40.7 Kj/mole=(87.3/18)*40.7 Kj =197.4 Kj

4. As per hess law, regardless of the no of steps in a chemical reaction, the total enthalpy change is sum of enthalpy changes of individual steps. Accordingly, the total enthaly= enthalpy required for converting water from 25 deg.c to 100 deg,c liquid + latent heat at 100 deg.c = 27.369+197.4 =224.769 Kj

5. Given Pbs(s)+1.5O2(g) ----->PbO(s)+ SO2(g), deltaH= -415.4 Kj (1)

PbO(s)+ C(s) ---------> Pb(s)+ Co(g), deltaH= 108.5 Kj (2)

Eq.1+Eq.2 gives Pbs(s)+C(s)+1.5O2------>Pb+SO2(g)+CO(g) deltaH= -415.4+108.5 Kj=-306.9 Kj

6. given 2H2(g)+ O2(g) ----->2H2O(l), deltaH= -573.6 Kj   (1)

N2O5(g)+ H2O(l)---------->2HNO3(l), deltaH= -73.7 Kj   (2), Eq,2*2 = 2N2O5(g)+ 2H2O(l) -----> 4HNO3(l), deltaH= -2*73.7= -147.4 Kj (2A)

Eq.1 +Eq.2A   = 2H2(g)+ O2(g)+ 2N2O5(g)-------> 4HNO3(l), deltaH=-147.4-573.6 KJ=-721Kj   (2B)

Eq. 3 is 0.5N2(g)+1.5O2(g)+0.5H2(g)------->HNO3, deltaH= -174.1 Kj (3)

Reversing it and mutlplying with 4 gives   4HNO3(l)-------> 2N2+ 6O2+2H2, deltaH= 4*174.1= 696.4 Kj (3A)

Addition of Eq.3A abd Eq.2B gives    2N2O5-----> 5O2+ 2N2, deltaH=-721+696.4 =-24.6 Kj (5)

Reversing the reaction gives 2N2+5O2------>2N2O5, deltaH= 24.6 Kj

dividing by 2 givees N2+2.5O2------->N2O5, deltaH= 24.6/2= 12.3 Kj    the desired reaction.

and 1/2N2(g)+3/2O2(g) +1/2H2(g) ------>HNO3(l), deltaH=-174.1 Kj (3)

4*eq.3 gives 2N2+ 6O2+3H2--->4HNO3(l) deltaH= -174.1*4 Kj=-696.4 Kj , reversing the reaction gives

4HNO3(l) ----->2N2(g)+ 6O2(g)+ 3H2(g), deltaH= 696.4 KJ (4

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