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Given That Δ\Deltaf for NH3 = -16

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Given that Δ\Deltaf for NH3 = -16.67 kJ/mol, calculate the equilibrium constant for the following reaction at 298 K. N2(g) + 3H2(g)  Given that  \Delta G°<sub>f</sub> for NH<sub>3</sub> = -16.67 kJ/mol, calculate the equilibrium constant for the following reaction at 298 K. N<sub>2</sub>(g)  + 3H<sub>2</sub>(g)    2NH<sub>3</sub>(g)  A)  1.20 * 10<sup>-</sup><sup>3</sup> B)  6.98 * 10<sup>5</sup> C)  8.36 * 10<sup>-</sup><sup>2</sup> D)  8.36 * 10<sup>2</sup> E)  1.42 * 10<sup>-</sup><sup>6</sup>  2NH3(g)


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