The following reaction is performed at 298 K ? 2 NO ( g )+ O 2( g ) leftharpoons 2 NO 2( g ) The standard free energy of formation of NO ( g ) is 86.6 kJ / mol at 298 K . What is the standard free energy of formation of NO 2( g ) at 298 K ?( K n=1.6 × 1012)

Question

The following reaction is performed at 298 K ? 2 NO ( g )+ O 2( g ) leftharpoons 2 NO 2( g ) The standard free energy of formation of NO ( g ) is 86.6 kJ / mol at 298 K . What is the standard free energy of formation of NO 2( g ) at 298 K ?( K n=1.6 × 1012) (A) R (298) in (1.6 × 1012) - 86600 (B) 86600 + R(298) in (1.6 × 1012) (C) 86600 - (In (1.6 × 1012)/R(298)) (D) 0.5[2 × 86600 - R(298) In (1.6 × 1012)]

Tardigrade Admin · Accepted Answer

The relationship between the standard Gibbs free energy change for reaction and the equilibrium constant is as shown below.

Δ G_{r x n}^{0} = - RT ln K_{p} \dots\dots (1)

The standard Gibbs free energy change for the reaction is given by the expression

Δ G_{r x n}^{0} = 2Δ G_{f}^{0} (N O_{2}) - 2Δ G_{f}^{0} (NO) \dots\dots .. (2)

From equations (1) and (2).

2Δ G_{f}^{0} (N O_{2}) - 2Δ G_{f}^{0} (NO) = - RT ln K_{p}

Δ G_{f}^{0} (N O_{2}) = 0.5 \times [2Δ G_{f}^{0} (NO) - RT ln K_{p}]

Substitute values in the above expression.

Δ G_{f}^{0} (N O_{2}) = 0.5 \times [2 \times 86600 - R (298) ln 1.6 \times 1 0^{12}] J / m o l

Q. The following reaction is performed at $298 K ?$
$2 NO (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$
The standard free energy of formation of $NO (g)$ is $86.6 k J / m o l$ at $298 K .$ What is the standard free energy of formation of $N O_{2} (g)$ at $298 K ? (K_{n} = 1.6 \times 1 0^{12})$

$R (298) in (1.6 \times 1 0^{12}) - 86600$

$86600 + R (298) in (1.6 \times 1 0^{12})$

$86600 - \frac{I n ( 1.6 \times 1 0 ^{12} )}{R ( 298 )}$

$0.5 [2 \times 86600 - R (298) I n (1.6 \times 1 0^{12})]$

Q. The following reaction is performed at 298K? 2NO(g)+O2​(g)⇌2NO2​(g) The standard free energy of formation of NO(g) is 86.6kJ/mol at 298K. What is the standard free energy of formation of NO2​(g) at 298K?(Kn​=1.6×1012)

R(298)in(1.6×1012)−86600

86600+R(298)in(1.6×1012)

86600−R(298)In(1.6×1012)​

0.5[2×86600−R(298)In(1.6×1012)]

Q. The following reaction is performed at $298 K ?$
$2 NO (g) + O_{2} (g) ⇌ 2 N O_{2} (g)$
The standard free energy of formation of $NO (g)$ is $86.6 k J / m o l$ at $298 K .$ What is the standard free energy of formation of $N O_{2} (g)$ at $298 K ? (K_{n} = 1.6 \times 1 0^{12})$

$R (298) in (1.6 \times 1 0^{12}) - 86600$

$86600 + R (298) in (1.6 \times 1 0^{12})$

$86600 - \frac{I n ( 1.6 \times 1 0 ^{12} )}{R ( 298 )}$

$0.5 [2 \times 86600 - R (298) I n (1.6 \times 1 0^{12})]$