How to practice concept:equilibrium-constant

Attempt a question, record the concept and assumptions, then check the available answer. Rework mistakes after revising the relevant concepts. Topic tags can overlap; review the question context.

Questions 1–4 of 4. Imported answers require independent review unless marked reviewed.

Topic practice 4 questionsPage 1 of 1

Question 34

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MTNAT1 markNeeds review
Consider the gas phase reaction:

At equilibrium for a particular temperature, the partial pressures of CO, O

, and CO

are found to be

atm,

atm, and 16 atm, respectively. The equilibrium constant for the reaction is _____

(rounded off to one decimal place).

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Question 51

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MTNAT2 marksNeeds review
Molten steel at 1900 K having dissolved hydrogen needs to be vacuum degassed.
The equilibrium partial pressure of hydrogen to be maintained to achieve 1 ppm (mass basis) of dissolved hydrogen is _____ Torr (rounded off to two decimal places).
Given: For the hydrogen dissolution reaction in molten steel (½ H

(g) = [H]), the equilibrium constant (expressed in terms of ppm of dissolved H) is:

1 atm = 760 Torr

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Question 35

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MTMCQ2 marksNeeds review
The equilibrium constant for the following reaction at 300 K is _____________.
C (graphite) + 2H₂ (g) → CH₄ (g)
Given: At 300 K, ΔH° = −74,900 J.mol⁻¹; ΔS° = −80 J.mol⁻¹.K⁻¹; R = 8.314 J.mol⁻¹.K⁻¹.
Your answerChoose one option.

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Question 37

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MTNAT2 marksNeeds review
In primary steelmaking, dissolved oxygen (O) reacts with carbon (C) to produce CO (g), at 1 atm pressure according to the reaction: C + O = CO (g). The equilibrium constant for this reaction is

, where T is in Kelvin. Assuming Henrian activity coefficient of both O and C to be unity, the dissolved oxygen content (in wt.%) of plain carbon steel melt with 0.7 wt.% C at 1600°C is __________ (answer up to four decimal places)

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