For the reaction mathrmN_2mathrmO_4(mathrmg) rightleftharpoons 2mathrmNO_2(mathrmg) mathrmK_p = 0.492 atm at 300mathrmK . mathrmK_c for the reaction at same temperature is \_\_\_\_\_\_ times 10^-2 . (textGiven: R = 0.082 text L atm mathrmmol^-1 textK^-1)

Numerical Answer Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

### Related Formula K_p = K_c cdot (RT)^Delta n_g ### Core Logic For the given gaseous equilibrium reaction: N_2O_4(g) rightleftharpoons 2NO_2(g) First, find the change in the number of moles of gas (Delta n_g): Delta n_g = n_p - n_r = 2 - 1 = 1 ### Step 1: Calculation Substitute the given values into the K_p - K_c relationship: K_p = 0.492 R = 0.082 T = 300text K 0.492 = K_c cdot (0.082 times 300)^1 K_c = frac0.4920.082 times 300 K_c = frac0.49224.6 K_c = 0.02 Converting to the requested format (x times 10^-2): K_c = 2 times 10^-2 So, the value is 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

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Equilibrium

More 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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