Question

JEE MAINS CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

Explanation:

i think the answer should be option C ....

Answer 2

Question:

For an equilibrium reaction N₂(g) + 3H₂(g) ⇄ 2 NH₃(g), $K_c$ = 64. What is the equilibrium constant for the reaction NH₃(g) ⇄ 1/2 N₂(g) + 3/2 H₂ (g)

Answer:

$\bigstar{\bold{Option\:1:\dfrac{1}{8} }}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• $K_c$ for the reaction N₂(g) + 3H₂(g) ⇄ 2 NH₃(g) is 64

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Equilibrium constant for the reaction NH₃(g) ⇄ 1/2 N₂(g) + 3/2 H₂

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The equilibrium constant for the given reaction N₂(g) + 3H₂(g) ⇄ 2 NH₃(g)

 is

   $K_c=\dfrac{[NH_3]^{2} }{[N_2][H_2]^{3} } = 64$

→ Let the equilibrium constant for the reaction NH₃(g) ⇄ 1/2 N₂(g) + 3/2 H₂ be $K_{c1}$

→ Hence equilibrium constant is given by

  $K_{c1} = \dfrac{[N_2]^{\frac{1}{2} } [H_2]^{\frac{3}{2} } }{[NH_3]}$

 $\dfrac{1}{K_{c1}}=\dfrac{[NH_3] }{[N_2]^{\frac{1}{2} } [H_2]^{\frac{3}{2} } }$

$\dfrac{1}{(K_{c1})^{2} }=\dfrac{[NH_3]^{2} }{[N_2]^{2 } [H_2]^{3 } }$

→ The RHS of the equation = $K_c$

→ Hence

  $\dfrac{1}{(K_{c1})^{2} } =K_c$

  $K_{c1}=\dfrac{1}{\sqrt{K_c} }$

→ Substituting value of $K_c$

  $K_{c1} = \dfrac{1}{\sqrt{64} } =\dfrac{1}{8}$

→ Hence option 1 is correct

$\Large{\underline{\underline{\bf{Notes:}}}}$

The ratio of the concentration of the products raised to the power of their stochiometric coefficients to the concentration of reactants raised to the power of their stochiometric coefficients at equilibrium state is called equilibrium constant.

aA + bB ⇄ cC + dD

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