0.3 g of ethane undergoes combustion at $27^{\circ} \mathrm{C}$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $0.5^{\circ} \mathrm{C}$. The heat evolved during combustion of ethane at constant pressure is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer) [Given : The heat capacity of the calorimeter system is $20 \mathrm{~kJ} \mathrm{~K}^{-1}, \mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Assume ideal gas behaviour. Atomic mass of C and H are 12 and $1 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively]

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Chemical Thermodynamics

0.3 g of ethane undergoes combustion at

2 7^{\circ} C

in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by

0. 5^{\circ} C

. The heat evolved during combustion of ethane at constant pressure is

____

kJ mol^{- 1}

.
(Nearest integer)
[Given : The heat capacity of the calorimeter system is

20 kJ K^{- 1}, R = 8.3 JK^{- 1} mol^{- 1}

.

Assume ideal gas behaviour.
Atomic mass of C and H are 12 and

1 g mol^{- 1}

respectively]

Answer: 1006

The amount of ethane is

\frac{0.3}{30} = 0.01

mol. In the bomb the heat released at constant volume is

q_{V} = C Δ T = 20 \times 0.5 = 10

kJ for this amount, giving

Δ U = - \frac{10}{0.01} = - 1000

kJ per mole. For the combustion

C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O (l)

the change in gaseous moles is

Δ n_{g} = 2 - \frac{9}{2} = - 2.5

. Then

Δ H = Δ U + Δ n_{g} R T = - 1000 + (- 2.5) (8.3 \times 1 0^{- 3}) (300) = - 1006.2

kJ mol

^{- 1}

. The heat evolved at constant pressure is about

1006

kJ mol

^{- 1}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer