Question

What will be the equilibrium constant at 127°C. If equilibrium constant at 27°C is 4 for the reaction N2+3H2⇌2NH3; ∆H=-46.06 kJ(1) 4×10-2(2) 2×10-3(3) 102(4) 4×102

Answer 1

The question is incomplete.

However.

Given:

Temperature = 27 degree

Equilibrium constant = 4 degree

N2 + 3H2 -> 2NH3

ΔH = -46.06 kJ

To find:

The equilibrium constant at 127 degree

Solution:

By Vant Hoff's equation,

ln ( k2 / k1 ) = Δ H / R ( 1 / T2 - 1 / T1 )

Here,

k1, k2 - Equillibrium constant at 27 and 127

T1, T2 - Temperatues

R = 8.314 J / Mol

Converting,

Celsius to kelvin

Substituting,

We get,

ln ( 4 ) - ln ( at 400 degree) = -46.06 * 10^3 / 8.314 ( 1 / 400 - 1 / 300 )

Solving,

Equilibrium constant at 400 k = 4 * 10^-2

Hence, the equilibrium constant at 127 degree is 4 * 10^-2

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