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  • The amount of heat liberated when 1.22 g of benzoic acid is burnt in bomb calorimeter at 298 K increases the temperature of 19 Kg of water by 1K. If the specific heat of water at 298 K is 1 ca

The amount of heat liberated when 1.22 g of benzoic acid is burnt in bomb calorimeter at 298 K increases the temperature of 19 Kg of water by 1K. If the specific heat of water at 298 K is 1 cal/g-degC. The value of heat of combustion (in kcal) of benzoic acid will be?  (Given Molar mass of benzoic acid = 122 gm-mole)

Option: 1

1220


Option: 2

1560


Option: 3

1860


Option: 4

1900


Answers (1)

best_answer

We have given,

Weight of Benzoic acid = 1.22 g

weight of water (m)  = 19 Kg = 19000 g

Increase in temp = 1K

Specific heat of water = 1 cal/g-degC

Heat liberated by benzoic acid, Q = ms\DeltaT

                                                      =  19000*1*1

                                                     = 19000 cal

1.22 g of benzoic acid liberates 19000 cal of heat

122 g of benzoic acid will liberate = (19000*122)/1.22

                                                      = 1900 kCal

Thus, the \mathrm{\Delta E} for the reaction is -1900 kCal

For calculation of \mathrm{\Delta H},  we have to write the reaction to find the \mathrm{\Delta n_g}:

\mathrm{C_6H_5COOH(s) + \frac{15}{2} O_2(g) \longrightarrow 7 CO_2(g) + 3 H_2O(l), \Delta n_g =-0.5}

\therefore \mathrm{\Delta H = \Delta E + (\Delta n_g) RT =-1900-0.5 \times 2 \times 298 \times 10^{-3}\approx -1900\ kCal}

 

Posted by

Shailly goel

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