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?

Missing in your question: ΔH = - 92.22 KJ. mol^-1 and the reaction equation is:

N2 (g) + 3H2 (g) ⇄2NH3 (g)

So according to this formula:

㏑ (K2/K1) = - ΔH/R (1/T2 - 1/T1)

when we have ΔH = 92.22 KJ. mol^-1 & K2 = X & K1 = 6.8X10^5 & T1 = 298 K &

T2=375 K & R constant = 8.314

So by substitution we can get the value of K2

㏑ (X/6.8x10^5) = 92.22/8.314 (1/375 - 1/298)

∴X = 6.75X10^5

∴K2 = 6.75X10^5

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

T ₂ = 375 K

Δ H = - 92,22 kJ/mol = - 92220 J/mol.

R = 8,314 J/K ·mol.

K ₁ = 6,8·10⁵.

K ₂ = ? 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.