You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Reaction Quotient

The free ener...

Question

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

8.4 \times 10^{- 5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

7.1 \times 10^{- 9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.2 \times 10^{- 10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.7 \times 10^{- 19}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $8.4 \times 10^{- 5}$
The relation between Gibbs free energy and K is given by
$- Δ G^{\circ} = 2.303 R T l o g K$
simplifying and taking antilog on both sides we get,
$K = anti log [\frac{- Δ G^{\circ}}{2.303 R T}] Putting the values we get, [\frac{- 78 \times 1000}{2.303 \times 8.314 \times 1000}] = 8.4 \times 10^{- 5}$

Suggest Corrections

Similar questions

Q. The free energy of the formation of

^{'} N O^{'}

78 k J / m o l

at the temperature of an automobile engine

(1000 K)

. What is the

K_{C}

equilibrium constant for this reaction at

1000 K

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ N O (g)

Q. The free energy of formation of NO is

78 k J m o l^{- 1}

at the temperature of

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) ⇌ N O (g)

Q. Calculate:

Δ G^{0}

b) the equilibrium constant for the formation of

N O_{2}

from NO and

O_{2}

at 298 K.

N O (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)

where

Δ_{f} G^{0} (N O_{2}) = 52.0

kJ/mol

Δ_{f} G^{0} (N O) = 87.0

kJ/mol

Δ_{f} G^{0} (O_{2}) = 0

kJ/mol.

Q. With the following information, determine standard Gibb's free energy of formation of

N_{2} O_{4}

(g).
(i)

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) \to N O (g)

Δ G^{0}

= 86.6 kJ
(ii)

N O (g) + \frac{1}{2} O_{2} (g) \to \frac{1}{2} N_{2} (g)

Δ G^{0}

= 34.82 kJ
(iii)

2 N O_{2} (g) \to N_{2} O_{4} (g)

Δ G^{0}

= 5.77 kJ

Q. The equilibrium constant

(K_{c})

for the reaction

N_{2} (g) + O_{2} (g) \to 2

(g)

at temperature

T

4 \times 10^{- 4}

. The value of

K_{c}

for the reaction,

N O (g) \to \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)

at the same temperature is:

Join BYJU'S Learning Program

Explore more

Reaction Quotient

Standard XII Chemistry

Join BYJU'S Learning Program

The free energy of formation of NO is 78 kJ mol−1 at the temperature of 1000 K. What is the equilibrium constant for this reaction at 1000 K? 12N2(g)+12O2(g)⇌NO(g)

The correct option is A 8.4×10−5The relation between Gibbs free energy and K is given by −ΔG∘=2.303 RT logK simplifying and taking antilog on both sides we get, K= anti log [−ΔG∘2.303RT] Putting the values we get,[−78×10002.303×8.314×1000]=8.4×10−5

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