What is the amount of work done when 0.5 mole of methane, CH4 (g), is subjected to combustion at 300 K ? (given, R = 8.314 J K-1 mol-1)

Question

What is the amount of work done when 0.5 mole of methane, CH4 (g), is subjected to combustion at 300 K ? (given, R = 8.314 J K-1 mol-1) (A) - 2494 J (B) - 4988 J (C) + 4988 J (D) + 2494 J

Tardigrade Admin · Accepted Answer

0.5 CH 4(g)+ O 2(g) arrow 0.5 CO 2(g)+ H 2 O (l) Δ n=0.5-1.5=-1 Work done =-Δ n R T=-(-1)* 8.314 × 30=2494 J

Q. What is the amount of work done when $0.5$ mole of methane, $C H_{4 (g)}$ , is subjected to combustion at $300 K$ ? (given, $R = 8.314 J K^{- 1} m o l^{- 1}$ )

$- 2494 J$

$- 4988 J$

$+ 4988 J$

$+ 2494 J$

$0.5 C H_{4} (g) + O_{2} (g) \to 0.5 C O_{2} (g) + H_{2} O (l)$

$Δ n = 0.5 - 1.5 = - 1$

Work done

$= - Δ n RT = - (- 1)^{*} 8.314 \times 30 = 2494 J$

Q. What is the amount of work done when 0.5 mole of methane, CH4(g)​, is subjected to combustion at 300K ? (given, R=8.314JK−1mol−1)

−2494J

−4988J

+4988J

+2494J