Quinone, which is used in the dye industry Can you please answer both, thanks! C
ID: 528560 • Letter: Q
Question
Quinone, which is used in the dye industry
Can you please answer both, thanks!
Calculate Delta H degree_f for B_5H_9 in kJ/mole given the following thermochemical equation f for and heat of formation data. 2 B_5H_9(g) + 12 O_2(g) rightarrow 5 B_2O_3(g) + 9 H_2O(g); Delta H = -8687.3 kJ Quinone, which is used in the dye industry and in photography, is an organic compound containing only C, H, and O. What is the empirical formula of the compound if you find that 0.105 g of the compound gives 0.257 g of CO_2 and 0.0350 g of H_2O when burned completely? Given a molecular weight of approximately 108 g/mol, what is its molecular formula?Explanation / Answer
1.
Enthalpy data : H2O(g)=-241.82 KJ, B2H3(g)= -843.8 Kj, O2=0
Enthalpy of reaction = Enthalpy of products- enthalpy of reactants
= 5* enthalpy of B2O3+ 9* enthalpy of H2O(g) – {2* enthalpy of B5H9(g) +12* enthalpy of formation of O2}
5, 9, 2 and 12 are coefficients of B2O3, H2O, B5H9 and O2 respectively in the reaction.
=5*-843.8 +9*(-241.8) –{ 2* enthalpy of B2O5(g)}= -8687.3
=-6395.2 -2* enthalpy of B2O5 =-8687.3
2* enthalpy of B2O5= 8687.3-6395.2= 2291.1
Enthalpy of B2O5= 2291.1/2= 1145.55 KJ/mole
2.
Moles= mass/molar mass
Molar masses : CO2= 44, H2O=18
Moles of prodjucts : CO2= 0.257/44=0.005841, moles of water= 0.0350/18= 0.001944
The combustion of C and H can be written as
C+O2-----àCO2 and H2+0.5O2-----àH2O
1 mole of CO2 required 1 mole of C and 1 mole of H2O requires 2 moles of H
Hence 0.005841 moles of CO2 required 0.005841 moles of C and 0.001944 moles of H2O requires 0.001944*2= 0.003889 moles of H
Mass of C= moles* atomic mass= 0.005841*12=0.070091 gm and mass of H= 0.003899*1=0.003899 gm
Total mass of quinone = 0.105, mass of O= total mass- mass of Carbon-mass of oxygen = 0.105-0.070091-0.003899 =0.03102
Moles of Oxygen atom =0.03102/16= 0.001939
Molar ratio of C: H:O= 0.005841 :0.003889:0.001939
Dividing by the slowest no 0.001939
The ratio becones 3 : 2:1
Empirical formula is C3H2O
Molar mass of empirical formula is 3*12+2+16= 54
Empirical formula* n= 108
Where n= no of repeating units
54*n=108
Hecne n= 108/54=2
The molecular formula is C6H4O2
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