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

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 2

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.