First slide

Thermodynamic properties

Question

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

Difficult

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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