show work please help 7. Use standard enthalpies of formation to determine Ho fo
ID: 510039 • Letter: S
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show work please help 7. Use standard enthalpies of formation to determine Ho for the reaction 2 Fe(s) FT3 CO2(g) CO(g) (Hr: CO2 (g) -393.5 kJ; Fe20,(s) 824.2 kJ; Coog) 110.5) A) 541.2 kJ B) -2336 kJ C) 541.2 kJ D) -24.8 kJ E) 24.8 kJ 8. Consider the following reaction: 6 CCs, graphite) 6 H20g) 3 O20g) C6H12O6(s) The heat of reaction for this represents the: A) heat of combustion of carbon D) standard heat of formation of glucose. E) none of the B) heat of formation of glucose. above. C) standard heat of combustion of carbon. Which of these aqueous solutions has the lowest boiling point? A) 1.25 M C6H12O B) 1.25 M KNO C) 1.25 D) None of the above E) Insufficient informationExplanation / Answer
Q.7 : Applying the Hess law of constant heat summation
Step.1: write all balanced equations.
(i) C + O2 -------> CO2 Hf1 = -393.5 kJ
(ii) C + 1/2O2 -------> CO Hf2 = -110.5 kJ
(iii) 2Fe + 3/2O2 ------> Fe2O3 Hf3 = -824.2 kJ
(iv) 2Fe + 3CO2 ----------> Fe2O3 + 3CO Hrxn = ?
Step.2 Change the stoichiometry according to equ (iv) and flip the equations if necessary to cancel terms on opposite sides.
(i) 3CO2 -------> 3C + 3O2 Hf1 = 3 x (+ 393.5) = +1180.5 kJ
(ii) 3C + 3/2O2 ---------> 3CO Hf2 = 3 x (-110.5) = -331.5 kJ
(iii) 2Fe + 3/2O2 ------> Fe2O3 Hf3 = -824.2 kJ
Step.3 Adding all the equations and Hf values and cancel the similar terms on opposite side (3C , 3O2 will cancel out )
2Fe + 3CO2 ----------> Fe2O3 + 3CO Hrxn = Hf1 + Hf2 +Hf3 = +1180.5 kJ--331.5 kJ-824.2 kJ = +24.8 kJ
ANS: E) +24.8 kJ
Q.8 : Given reaction is
6C + 6H2 + 3O2 -------------> C6H12O6
The standard enthalpy of this reaction represents heat of formation of glucose from its constituent elements.
Q.9 :
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