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) mathrm~N2( mathrm~g)+(1/2) mathrmO2( mathrm~g) arrow mathrmNO( mathrmg) (Anti log 0.9263 = 8.459)

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) mathrm~N2( mathrm~g)+(1/2) mathrmO2( mathrm~g) arrow mathrmNO( mathrmg) (Anti log 0.9263 = 8.459) (A) 8.4× 10- 5 (B) 7.1× 10- 9 (C) 4.2× 10- 10 (D) 1.7× 10- 19

Tardigrade Admin · Accepted Answer

Ke q=antilog((- Δ Go/2.303 R T)) Δ Go=78 kJ/mol=78000 J/mol R=8.314 JK- 1mol- 1,T=1000 K Putting these values in eqn. (i) we get, Ke q=antilog[(- 78000/2.303 × 8.314 × 1000)] Ke q=antilog(.-4.0737.) or Ke q=8.4× 10- 5

Q. 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?

$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) \to NO (g)$

(Anti log 0.9263 = 8.459)

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

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

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

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

Q. 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? 21​ N2​( g)+21​O2​( g)→NO(g) (Anti log 0.9263 = 8.459)

8.4×10−5

7.1×10−9

4.2×10−10

1.7×10−19

Keq​=antilog(2.303RT−ΔGo​) ΔGo=78kJ/mol=78000J/mol R=8.314JK−1mol−1,T=1000K Putting these values in eqn. (i) we get, Keq​=antilog[2.303×8.314×1000−78000​] Keq​=antilog(−4.0737) or Keq​=8.4×10−5

Q. 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?

$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) \to NO (g)$

(Anti log 0.9263 = 8.459)

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

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

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

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