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**Problem**: Consider the following reaction at 800. K:N2(g) + 3F2(g) → 2NF3(g)An equilibrium mixture contains the following partial pressures: PN2 = 0.021 atm, PF2 = 0.063 atm, PNF3 = 0.48 atm.Calculate ΔG° for the reaction at 800. K.

###### FREE Expert Solution

Recall that the ** equilibrium constant** is the ratio of the products and reactants.

We use **K _{p}** when dealing with pressure.

The ** K_{p} expression** for the reaction is:

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{products}}}{{\mathbf{P}}_{\mathbf{reactants}}}{\mathbf{=}}\frac{{{\mathbf{P}}_{{\mathbf{NF}}_{\mathbf{3}}}}^{\mathbf{2}}}{{\mathbf{P}}_{{\mathbf{N}}_{\mathbf{2}}}\mathbf{}{{\mathbf{P}}_{{\mathbf{F}}_{\mathbf{2}}}}^{\mathbf{3}}}}$

We’re given the following equilibrium pressures:

###### Problem Details

Consider the following reaction at 800. K:

N_{2}(g) + 3F_{2}(g) → 2NF_{3}(g)

An equilibrium mixture contains the following partial pressures:

P_{N2} = 0.021 atm, P_{F2} = 0.063 atm, P_{NF}_{3} = 0.48 atm.

Calculate ΔG° for the reaction at 800. K.

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