Question

The equilibrium constant k for the synthesis of ammonia is 6.8x105 at 298 k. what will k be for the reaction at 375 k?

Answer 1

Answer is: K be for the reaction at 375 K is 326.
Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁ = 298 K
T₂ = 375 K
ΔH = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K·mol.
K₁ = 6,8·10⁵.
K₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.

Answer 2

The value of K for the reaction at 375 k is : 326

Given data :

Initial temperature ( T1 ) = 298 k

rate constant ( k1 ) = 6.8 * 10⁵

Final temperature ( T2 ) = 375 k

Determine the value of  K2  

applying the relationship below

Log ( K₂ / K₁ ) = ΔH / 2.303 * R  * ( T₂-T₁ / T₂T₁ ) ----- ( 1 )

equation ( 1 ) becomes

Log K₂ - log (6.8 * 10⁵   ) =  - 7100940 / 213967725

Log K₂ - ( 5 + log 6.8 ) = - 3.318

therefore Log K₂ = 2.5145

K₂ = 10^2.5145

    = 326

Hence we can conclude that The value of K for the reaction at 375 k is : 326

Learn more about equilibrium constant : https://brainly.com/question/12270624

Attached below is the complete question