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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
The reaction between ${N_2}$ and ${H_2}$ to form ammonia has ${K_c} = 6 \times {10^{ - 2}}$ at the temperature $500\,^oC$. The numerical value of ${K_p}$ for this reaction is
  • ✓
    $1.5 \times {10^{ - 5}}$
  • B
    $1.5 \times {10^5}$
  • C
    $1.5 \times {10^{ - 6}}$
  • D
    $1.5 \times {10^6}$
✓

Answer

Correct option: A.
$1.5 \times {10^{ - 5}}$
(a) ${K_p} = {K_c}{(RT)^{\Delta n}}$;   $\Delta n = 2 - 4 = - 2$

${K_p} = 6 \times {10^{ - 2}} \times {(0.0812 \times 773)^{ - 2}}$

${K_p} = \frac{{6 \times {{10}^{ - 2}}}}{{{{(0.0812 \times 773)}^2}}} = 1.5 \times {10^{ - 5}}$.

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