Question

Calculate the work done during the combustion of 0.138 kg of ethanol, C₂H₅OH(l) at 300 K.
Given: R = 8.314 Jk⁻¹ mol⁻¹, molar mass of ethanol = 46 g mol⁻¹.

पर्याय

• −7482 J

• 7482 J

• −2494 J

• 2494 J

Answer 1

7482 J

Explanation:

Combustion of ethanol is as:

\[\ce{C2\underset{\underset{1}{↓}}{H5}OH + 3\underset{\underset{3}{↓}}{O}_2 -> 2C\underset{\underset{2}{↓}}{O} + 3\underset{\underset{3}{↓}}{H2O}}\]

0.138 kg = 138 g = `138/46` = 3 mole

\[\ce{3C2H5OH(l) + 9O2(g) -> 6CO2(g) + 9H2O(l)}\]

Δn = 6 − 9 = −3

Work = −ΔnRT

= −(−3) × 8.314 × 300

=  7482 J