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6.1. Equilibrium - 1 (chemical Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.1. Equilibrium - 1 (chemical Equilibrium)1 Mark
MCQ
For the equilibrium ${N_2} + 3{H_2}$ $\rightleftharpoons$ $2N{H_3},{K_c}$ at $1000\,K$ is $2.37 \times {10^{ - 3}}$. If at equilibrium $[{N_2}] = 2\,M,\,[{H_2}] = 3\,M$, the concentration of $N{H_3}$ is
  • A
    $0.00358$
  • B
    $0.0358$
  • ✓
    $0.358$
  • D
    $3.58$
✓

Answer

Correct option: C.
$0.358$
(c) ${K_c} = \frac{{{{[N{H_3}]}^2}}}{{[{N_2}]\,\,{{[{H_2}]}^3}}}$

$2.37 \times {10^{ - 3}} = \frac{{{x^2}}}{{[2]\,\,{{[3]}^3}}} = {x^2} = 0.12798$

$x = 0.358 \,M$

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