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15 April, 01:26

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?

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  1. 15 April, 01:41
    0
    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.
  2. 15 April, 01:55
    0
    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
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