High School

1. For the reaction:

\[ 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g) \]

\[ \Delta H^\circ = -114.2 \text{ kJ} \quad \text{and} \quad \Delta S^\circ = -146.5 \text{ J/K} \]

The equilibrium constant for this reaction at 275.0 K is ______.

Assume that \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent of temperature.

2. For the reaction:

\[ \text{NH}_4\text{NO}_3(aq) \rightarrow \text{N}_2\text{O}(g) + 2\text{H}_2\text{O}(l) \]

\[ \Delta H^\circ = -149.6 \text{ kJ} \quad \text{and} \quad \Delta S^\circ = 99.9 \text{ J/K} \]

The equilibrium constant for this reaction at 256.0 K is ______.

Assume that \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent of temperature.

Answer :

1. For the reaction

2NO(g) + O2(g)2NO2(g)

Kc = 6.64 × 10^21

2. For the reaction

NH4NO3(aq)N2O(g) + 2H2O(l)

Kc = 2.17 × 10^29.

1. For the reaction

2NO(g) + O2(g)2NO2(g)

H° = -114.2 kJ and S° = -146.5 J/K.

The equilibrium constant for this reaction at 275.0 K is given by:

Kc = e^(-ΔG/RT)

where R = 8.314 J/mol.K;

T = 275 K;

ΔH° = -114.2 kJ/mol and ΔS° = -146.5 J/mol.K

First, ΔG° = ΔH° - TΔS°

ΔG° = -114200 - 275 (-146.5/1000)Δ

G° = -114200 + 40.2125

ΔG° = -114159.79 J/mol

Kc = e^(-ΔG°/RT)

Kc = e^(-(-114159.79)/(8.314 × 275))

Kc = e^(49.51)

Kc = 6.64 × 10^21

2. For the reaction

NH4NO3(aq)N2O(g) + 2H2O(l)

H° = -149.6 kJ and S° = 99.9 J/K.

The equilibrium constant for this reaction at 256.0 K is given by:

Kc = e^(-ΔG/RT)

where R = 8.314 J/mol.K; T = 256 K;

ΔH° = -149.6 kJ/mol and ΔS° = 99.9 J/mol.K

First, ΔG° = ΔH° - TΔS°

ΔG° = -149600 - 256 (99.9/1000)

ΔG° = -149600 + 25.554

ΔG° = -149574.446 J/mol

Kc = e^(-ΔG°/RT)

Kc = e^(-(-149574.446)/(8.314 × 256))

Kc = e^(68.153)

Kc = 2.17 × 10^29.

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1. The equilibrium constant for the reaction 2NO(g) + O₂(g) → 2NO₂(g) at 275.0 K is 840.562.

2. The equilibrium constant for the reaction NH₄NO₃(aq) → N₂O(g) + 2H₂O(l) at 256.0 K is 194.43.

1. To calculate equilibrium constant for the reaction 2NO(g) + O₂(g) → 2NO₂(g) at 275.0 K, we use the following equation:

ΔG° = -RTlnK,

where

ΔG° = -114.2 kJ/mol - (275.0 K)(-146.5 J/K)(1 kJ/1000 J)

= -114.2 kJ/mol + 40.2475 kJ/mol

= -73.95 kJ/mol

R = 8.314 J/mol·K

T = 275.0 K

Substituting these values into the equation above gives:

ΔG° = -RTlnK

-73.95 kJ/mol = -(8.314 J/mol·K)(275.0 K)lnK

lnK = 6.7156K = e6.7156K = 840.56

Thus, the equilibrium constant for the reaction at 275.0 K is 840.562.

2. To calculate the equilibrium constant for the reaction NH₄NO₃(aq) → N₂O(g) + 2H₂O(l) at 256.0 K, we can use the following equation:

ΔG° = -RTlnK,

where

ΔG° = -149.6 kJ/mol - (256.0 K)(99.9 J/K)(1 kJ/1000 J)

= -149.6 kJ/mol + 25.5264 kJ/mol

= -124.07 kJ/mol

R = 8.314 J/mol·K

T = 256.0 K

Substituting these values into the equation above gives:

ΔG° = -RTlnK

-124.07 kJ/mol = -(8.314 J/mol·K)(256.0 K)lnK

lnK = 5.2656K = e5.2656K = 194.43

Thus, the equilibrium constant for the reaction at 256.0 K is 194.43.

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