JEE Main · 2006 · 30 Apr 2006 · Offline · Chemical Thermodynamics

(Δ H - Δ U)

for the formation of carbon monoxide (CO) from its elements at 298 K is
(

R = 8.314 J K^{- 1} mol^{- 1}

)

Answer: (A)

Solution

The formation of carbon monoxide is

C (s) + \frac{1}{2} O_{2} (g) \to CO (g)

, and the gap between the two energies is

Δ H - Δ U = Δ n_{g} R T

, counting only gaseous species. Here the gaseous moles change from

\frac{1}{2}

on the left to

1

on the right, so

Δ n_{g} = 1 - \frac{1}{2} = - \frac{1}{2}

. Thus

Δ H - Δ U = - \frac{1}{2} \times 8.314 \times 298 = - 1238.78 J mo l^{- 1}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option