The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500 K. N2=1.5×10-2M, H2=3.0×10-2M and [NH3] = 1.2 × 10 -2M. Calculate the equilibrium constant.

# The following concentrations were obtained for the formation of ${\mathrm{NH}}_{3}$ from ${\mathrm{N}}_{2}$ and ${\mathrm{H}}_{2}$ at equilibrium at 500 K.  and

1. A

2. B

3. C

4. D

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### Solution:

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[{\mathrm{NH}}_{3}\left(\mathrm{g}\right)\right]}^{2}}{\left[{\mathrm{N}}_{2}\left(\mathrm{g}\right)\right]{\left[{\mathrm{H}}_{2}\left(\mathrm{g}\right)\right]}^{3}}=\frac{{\left(1.2×{10}^{-2}\right)}^{2}}{\left(1.5×{10}^{-2}\right){\left(3.0×{10}^{-2}\right)}^{3}}$

= 3.55 $×{10}^{2}$

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