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Chemical Equilibrium question

2021 · 27 Jul · Shift 2 · Q13
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Chemical Equilibrium question

2021 · 27 Jul · Shift 2 · Q13

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The equilibrium constant for the reaction A(s) ⇌\rightleftharpoons⇌ M(s) +12{1 \over 2}21​ O2O_2O2​(g) is Kp = 4. At equilibrium, the partial pressure of O2O_2O2​ is ‾\underline{\hspace{2cm}}​ atm. (Round off to the nearest integer)
Numerical answer
View written solutionFree

Correct answer: 16

  1. Write the reaction and equilibrium expression

    The reaction is: A(s)⇌M(s)+12O2(g)A(s) \rightleftharpoons M(s) + \frac{1}{2}O_2(g)A(s)⇌M(s)+21​O2​(g)

    For equilibrium involving solids, the activities of pure solids are taken as 111.

    Therefore, Kp=(PO2)1/2K_p = \left(P_{O_2}\right)^{1/2}Kp​=(PO2​​)1/2

  2. Use the given value of KpK_pKp​

    Given: Kp=4K_p = 4Kp​=4

    So, 4=(PO2)1/24 = \left(P_{O_2}\right)^{1/2}4=(PO2​​)1/2

  3. Solve for PO2P_{O_2}PO2​​

    Squaring both sides, PO2=42=16 atmP_{O_2} = 4^2 = 16\,\text{atm}PO2​​=42=16atm

  4. Nearest integer

    16 atm16\,\text{atm}16atm is already an integer.

Final Answer: 16\boxed{16}16​

  1. Comparison with stored correct answer

    Stored correct answer = 161616

    My derived answer = 161616

    Hence, they agree.

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