Solve question 7:

7. The difference between heat of reaction at constant volume and constant pressure of the reaction.
${\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{5}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
At 300 K in Kcal is [Given: R=2 cal/mol. K]
(A) -0.3 Kcal             (B) +0.3 Kcal           (C) -0.9 Kcal        (D) None of these.

Dear student,
  

The given reaction is;  C₂H_2(g)  + $\frac{5}{2}$ O_2(g)  $\to$2 CO_2(g)  + H₂O_(l) 
Again we know that ; heat at constant volume = $∆$U ( Change in internal energy )
                                   heat at constant pressure = $∆$ H ( Change in enthalpy )
                                     Now from the first law of thermodynamics for the reactions; $∆$H = $∆$ U - $∆$n_g RT ........(i)
                    From the above relation $∆$n_g = number of moles of gaseous products - number of moles of gaseous reactants
                                                                       = 2 - ( $\frac{5}{2}$+ 1) =  - $\frac{3}{2}$
                       Now on putting the values inequation (i) ; $∆$H = $∆$ U - (- $\frac{3}{2}$$\times$2$\times$300 )
                                                            Then ; $∆$H - $∆$ U = 900 cal /mol or 0.9 kcal/mol
                       Again according to the question asked ; $∆$ U - $∆$H = - 0.9 kcal/mol (Option C)
Regards