The following concentrations were observed at 500mathrmK for the formation of mathrmNH_3 from mathbfN_2 and mathrmH_2. At equilibrium: [mathrmN_2] = 2times 10^-2mathrmM, [mathrmH_2] = 3times 10^-2mathrmM and [mathrmNH_3] = 1.5times 10^-2mathrmM. Equilibrium constant for the reaction is ________.

Numerical Answer Type:
Enter a numerical value Answer: 417 to 417 +4 marks

Solution & Explanation

### Related Formula mathrmN_2(g) + 3mathrmH_2(g) rightleftharpoons 2mathrmNH_3(g) K_c = frac[mathrmNH_3]^2, [mathrmN_2][mathrmH_2]^3 ### Core Logic Substituting the given equilibrium concentrations into the equilibrium constant expression: K_c = frac(1.5 times 10^-2)^2, (2 times 10^-2) times (3 times 10^-2)^3 Evaluating the values step-by-step: K_c = frac2.25 times 10^-4, (2 times 10^-2) times (27 times 10^-6) ### Step 1: Final Arithmetic Integration K_c = frac2.25 times 10^-4, 54 times 10^-8 = frac2.25, 54 times 10^4 = 0.041666 times 10^4 approx 416.67 Rounding to the nearest integer yields **417**. ### Pattern Recognition Pay close attention to the cubic exponent in the denominator derived from the hydrogen stoichiometric coefficient (3). Small calculation errors here can significantly alter the result. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium

Reference Study Guides

Chemical Equilibrium

More Chemical Equilibrium Previous-Year Questions — Page 6

Q62 jee_main_2024_31_jan_morning Equilibrium Constant
For the given reaction, choose the correct expression of K_C from the following :- Fe_(aq)^3+ + SCN_(aq)^- rightleftharpoons (FeSCN)_(aq)^2+
  • A. K_C = frac[FeSCN^2+][Fe^3+][SCN^-]
  • B. K_C = frac[Fe^3+][SCN^-][FeSCN^2+]
  • C. K_C = frac[FeSCN^2+][Fe^3+]^2[SCN^-]^2
  • D. K_C = frac[FeSCN^2+]^2[Fe^3+][SCN^-]

Solution

### Related Formula K_C = frac[textProducts][textReactants] ### Core Logic K_C = fractextProducts ion conc.textReactants ion conc. K_C = frac[FeSCN^2+][Fe^3+][SCN^-] ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

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