Plz post the sol...

Q109. At 300 K the standard enthalpies of formation of C6H5COOH(s), CO2 (g) and H2O(i) are – 408, – 393 and – 286 kJ mol–1 respectively. Calculate the heat of combustion of benzoic acid at contant volume :

    (1) + 3201 kJ

    (2) + 3199.75 kJ

    (3) – 3201 kJ

    (4) – 3199.75 kJ

Dear student,


C6H5COOH(s) +13/2 O2(g)  →7CO2(g)   +3H2O(l)  

q(p)  = ΔH = ΔH of products  -  ΔH of reactants

       =   7(−393) + 3(−286) – (−408) – 0


 
q(p) = −2751–858 + 408   = −3201kJ/mol
 
According to balance chemical reaction

Δn = number of gaseous product - number of  gaseous reactant molecules 
    
Δn =    7−13/2  = 0.5


q(V) = ΔE = ΔH – Δn x RT
  
 q(V)    =−3201 – 0.5 × 8.314 × 10−3× 300
   
  q(V) =−3202.25kJ/mol

Answer: q(V) =−3202.25kJ/mol ( option 3)  

Regards


 

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I don't whether the answer is correct but this should be the process..!!

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