Question

The equilibrium constant kp at 427°c for the reaction n 2( g) 3h 2( g) 2nh 3( g) is 9. 4 × 10 –5. what is δ g° for the reaction under these conditions? ( r = 8. 314 j/k • mol)

Answer 1

ΔG° for the reaction is 5.47kJ mol⁻¹.

The energy that a substance has available for utilization in a chemical reaction or transformation is known as the Gibbs free energy. Things frequently change into other things that have less Gibbs free energy. The Gibbs free energy change indicates whether a chemical reaction will take place spontaneously or not.

By using the formula;

ΔG° = −RTlnKp

Where,

R = 8.3Jk⁻¹mol⁻¹

T = Temperature = 427 + 273 = 700 K

Kp = 8×10⁻⁵(given)

Substituting the value, we get,

ΔG° = −8.3 × 700 × ln(23×10⁻⁵)

ΔG° = −8.3 × 700 × (ln(2³)+ln 10⁻⁵)

       =  - 8.3 × 700 × (ln(2³)+ln 10⁻⁵)

       = − 8.3 × 700 × (2.07−11.5)

       =5.47×10⁴Jmol¹

       =5.47kJ mol⁻¹

Therefore,  ΔG° for the reaction is 5.47kJ mol⁻¹.

Learn more about Gibbs free energy here:

brainly.com/question/13765848

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