NAME 9. Compute the molar enthalpy of combustion of glucose (CHo GI,206 (s) +602
ID: 557960 • Letter: N
Question
NAME 9. Compute the molar enthalpy of combustion of glucose (CHo GI,206 (s) +602 (g) 6CO2 (g) + 6H20(g) Given that combustion of 0.305 g of glucose caused a raise in temperature of 6. 30 Cof a cet lorimeter with total heat heat capacity C75 JC. Note that what is given is the beat capacity of the calorimeter, not the specithic heat capacity of the substance inside the calorimeter. The molecular weight of glucose is 180.2 amu. Spes Answer: 1 282x10, -kJ/mol 10. Given the following thermochemical data: pts 2P(s) + 3 Cl2(g) 2 PC, (i) =-636 kJ/mol AH = 138 klimol compute the enthalpy of formation of PCI, (8), that is the enthalpy change for the reaction PCs) +5/2 Cl2 (g) PCIS (s) A-? Answer: AH-456 kJ/molExplanation / Answer
Q9
mol = mass/MW = 0.305/180 = 0.001694
dT = +6.30
Cp = 755 J/C
Qtotal = Cp*dT = 755*6.30 = 4756.5 J
Hrxn = -Q/n = -4756.5 / (0.001694)
HRxn = -2807851.23 J/mol
HRxn 0 -2.81*10^3 kJ/mol
Q10.
Apply HEss law
divide rxn 1 by 2
P(s) + 3/2Cl2 = PCl3 H = 1/2*(-636) = -318
invet rxn 2
PCl3 + Cl2 = PCl5 H = -138
add all
PCl3 + Cl2 + P(s) + 3/2Cl2 = PCl3+PCl5 H = -318+ -138 = -456 kJ/mol
P(s) + 5/2Cl2 = PCl5 H = -456 kJ/mol
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