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Chemical Equilibrium question

2024 · 29 Jan · Shift 2 · Q24
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  5. /2024 · 29 Jan · Shift 2 · Q24

Chemical Equilibrium question

2024 · 29 Jan · Shift 2 · Q24

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The following concentrations were observed at 500 K500 \mathrm{~K}500 K for the formation of NH3\mathrm{NH}_3NH3​ from N2\mathrm{N}_2N2​ and H2\mathrm{H}_2H2​. At equilibrium ; [N2]=2×10−2M,[H2]=3×10−2M\left[\mathrm{N}_2\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right]=3 \times 10^{-2} \mathrm{M}[N2​]=2×10−2M,[H2​]=3×10−2M and [NH3]=1.5×10−2M\left[\mathrm{NH}_3\right]=1.5 \times 10^{-2} \mathrm{M}[NH3​]=1.5×10−2M. Equilibrium constant for the reaction is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 417

  1. Write the balanced reaction

For formation of ammonia:

N2+3H2⇌2NH3\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}N2​+3H2​⇌2NH3​
  1. Write the expression for equilibrium constant

In terms of concentration,

Kc=[NH3]2[N2][H2]3K_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}Kc​=[N2​][H2​]3[NH3​]2​
  1. Substitute the given equilibrium concentrations

Given:

[N2]=2×10−2 M[\mathrm{N_2}] = 2 \times 10^{-2} \,\mathrm{M}[N2​]=2×10−2M [H2]=3×10−2 M[\mathrm{H_2}] = 3 \times 10^{-2} \,\mathrm{M}[H2​]=3×10−2M [NH3]=1.5×10−2 M[\mathrm{NH_3}] = 1.5 \times 10^{-2} \,\mathrm{M}[NH3​]=1.5×10−2M

So,

Kc=(1.5×10−2)2(2×10−2)(3×10−2)3K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}Kc​=(2×10−2)(3×10−2)3(1.5×10−2)2​
  1. Calculate numerator and denominator

Numerator:

(1.5×10−2)2=2.25×10−4(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}(1.5×10−2)2=2.25×10−4

Denominator:

(3×10−2)3=27×10−6(3 \times 10^{-2})^3 = 27 \times 10^{-6}(3×10−2)3=27×10−6

Then,

(2×10−2)(27×10−6)=54×10−8=5.4×10−7(2 \times 10^{-2})(27 \times 10^{-6}) = 54 \times 10^{-8} = 5.4 \times 10^{-7}(2×10−2)(27×10−6)=54×10−8=5.4×10−7
  1. Find KcK_cKc​
Kc=2.25×10−45.4×10−7K_c = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}}Kc​=5.4×10−72.25×10−4​ Kc=2.255.4×103K_c = \frac{2.25}{5.4} \times 10^3Kc​=5.42.25​×103 Kc=0.4167×103=416.7K_c = 0.4167 \times 10^3 = 416.7Kc​=0.4167×103=416.7
  1. Final integer answer
Kc≈417K_c \approx 417Kc​≈417
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