Question

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

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Answer 1

Answer :

• Equilibrium constant, $\rm{K_c}$ = 1.06 × 10³.

Step-by-step explanation:

Given that,

• Equilibrium concentration of H₂ = 3.0 × 10⁻² M.
• Equilibrium concentration of N₂ = 1.5 × 10⁻² M.
• Equilibrium concentration of NH₃ = 1.2 × 10⁻² M.

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According to the question, the reaction will be:-

$\rm{N_2(g)+3H_2(g)\leftrightarrow{}2NH_3(g)}$

Also we know,

▪ [H₂] = 3.0 × 10⁻² M.

▪ [N₂] = 1.5 × 10⁻² M.

▪ [NH₃] = 1.2 × 10⁻² M.

Putting these values in the equation,

$\implies\rm{K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}}$

We get,

$\implies\rm{K_c=\dfrac{[NH_3(g)]^2}{[N_2(g)][H_2(g)]^3}}$

$\implies\rm{K_c=\dfrac{(1.2\times{}10^{-2})^2}{(1.5\times{}10^{-2})(3.0\times{}10^{-2})^3}}$

$\implies\rm{K_c=0.106\times{}10^4=}\:\bold{1.06\times{}10^3.}$