The free energy of formation of NO is 78 kJ mol -1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K? (1/2) N2(g)+(1/2) O2(g) N O(g)

Question

The free energy of formation of NO is 78 kJ mol -1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K? (1/2) N2(g)+(1/2) O2(g) N O(g) (A) 8.4 × 10-5 (B)  7.1× 10-9  (C)  4.2× 10-10  (D)  1.7× 10-19

Tardigrade Admin · Accepted Answer

Free energy and equilibrium constant are related as, Δ G=-2.303 R T log K 78 × 103=-2.303 × 8.314 × 1000 × log K log K=-(78/2.303 × 8.314) =-4.073 K = antilog (-4.0 .73) =8.43 × 10-5

Q. The free energy of formation of $NO$ is $78 k J m o l^{- 1}$ at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at $1000 K$ ?
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) NO (g)$

$8.4 \times 1 0^{- 5}$

$7.1 \times 1 0^{- 9}$

$4.2 \times 1 0^{- 10}$

$1.7 \times 1 0^{- 19}$

Free energy and equilibrium constant are related as,
$Δ G = - 2.303 RT lo g K$
$78 \times 1 0^{3} = - 2.303 \times 8.314 \times 1000 \times lo g K$
$lo g K = - \frac{78}{2.303 \times 8.314}$
$= - 4.073$
$K =$ antilog $(- 4.0.73)$
$= 8.43 \times 1 0^{- 5}$

Q. The free energy of formation of NO is 78kJmol−1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000K? 21​N2​(g)+21​O2​(g)NO(g)

8.4×10−5

7.1×10−9

4.2×10−10

1.7×10−19