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

Given :-

•[H₂] = 3.0 × 10⁻² M.

•[N₂] = 1.5 × 10⁻² M.

•[NH₃] = 1.2 × 10⁻² M.

To Find :-

• Equilibrium constant, $\rm{K_c}$

Solution :-

Given reaction,

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

$\underline{\:\textsf{ We know that :}}$

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

$\underline{\:\textsf{ Putting the given values :}}$

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

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

$\dashrightarrow \rm{K_c=0.106\times{}10^4}\:$

$\dashrightarrow {\boxed{\frak{\purple{K_c = 1.06 × 10³}}}}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{Equilibrium constant is </p><p>\textbf{1.06 × 10³ }}}.$

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Answer 2

The following concentrations were obtained for the formation of NH3from 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.