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Q.

Heat of combustion of benzoic acid at constant volume and 298 K is –3233kJ/mole. When 0.5g of benzoic acid is burnt in bomb calorimeter, the temperature of calorimeter increased by $0 . 53^{0} c$ . Now in the same bomb calorimeter when 1g of ethane is burnt then temperature increased by $2 . 04^{0} c$ . ΔH for combustion of ethane is

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a

–1530KJ/mole

b

–1536.2KJ/mole

c

–1522.8KJ/mole

d

+1536.2KJ/mole.

answer is B.

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Detailed Solution

\Delta U=Q\times\Delta t\times \frac{M}{m}

Q = heat capacity of the calorimeter
Δt = rise in temperature
M = Molar mass of substance
m = mass of substance taken

\large {(\Delta U)_{{C_6}{H_5}COOH}}\; = \;Q\; \times \;0.53\; \times \;\frac{{122}}{{0.5}}

\large {(\Delta U)_{{C_6}{H_6}}}\; = \;Q\; \times \;2.04\; \times \;\frac{{30}}{1}

Since bomb calorimeter is same, Q is same

\large \frac{{ - 3233}}{{{{(\Delta U)}_{{C_2}{H_6}}}}}\; = \;\frac{{0.53\; \times \;122}}{{2.04\; \times \;0.5\; \times \;30}}

\large \frac{{ - 3233}}{{{{(\Delta U)}_{{C_2}{H_6}}}}}\; = \;2.113

\large {(\Delta U)_{{C_2}{H_6}}}\; = \;\frac{{ - 3233}}{{2.113}}\; = \; - 1530KJ

\large \mathop {{C_2}{H_6}}\limits_{(g)} \;\; + \;\mathop {7/2{O_2}}\limits_{(g)} \; \to \;\mathop {2C{O_2}}\limits_{(g)} \; + \;\mathop {3{H_2}O}\limits_{(l)}

\large \Delta n_g\; = \;2 - \left( {1 + \frac{7}{2}} \right)\; = \;2 - \;\frac{9}{2}\; = \; - 5/2

ΔH= ΔU + ΔnRT

= -1530 + (-5/2) x 8.314 x 10-3 x 298
= -1536.2KJ/mole

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Heat of combustion of benzoic acid at constant volume and 298 K is –3233kJ/mole. When 0.5g of benzoic acid is burnt in bomb calorimeter, the temperature of calorimeter increased by 0.530c. Now in the same bomb calorimeter when 1g of ethane is burnt then temperature increased by 2.040c. ΔH for combustion of ethane is