At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO 2 in equilibrium with solid carbon has 90.55% CO by mass Calculate K c for this reaction at the above example. - Wired Faculty

Equilibrium

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At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
$straight C left parenthesis straight s right parenthesis space plus space CO subscript 2 left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space 2 CO left parenthesis straight g right parenthesis$
Calculate K_cfor this reaction at the above example.

Solution

If the total mass of the mixture of CO and CO₂ is 100g. then the mass of CO=90.55g and mass of
CO₂= 100-90.55=9-45g
therefore,
Number of moles of CO= 90.55/28=3.23
Number of moles of CO= 9.45/44 =0.22
$therefore comma space straight p subscript CO space equals fraction numerator 3.23 over denominator 3.23 space plus 0.22 end fraction space straight x space 1 space atm space equals space 0.94 space atm straight p subscript CO subscript 2 end subscript space equals space fraction numerator 0.22 over denominator 3.23 plus 0.22 end fraction space straight x space 1 space atm space equals 0.06 space atm straight K subscript straight p space equals space straight p squared subscript CO subscript 2 end subscript over straight p subscript CO space equals space fraction numerator left parenthesis 0.94 right parenthesis squared over denominator 0.06 space end fraction space equals 14.25 increment straight n subscript straight g space equals 2 minus 1 space equals 1 straight K subscript straight p space equals straight K subscript straight c left parenthesis RT right parenthesis straight K subscript straight c space equals space straight K subscript straight p over RT space equals fraction numerator 14.25 over denominator 0.08210 space straight x 1127 end fraction space equals 0.154$

Some More Questions From Equilibrium Chapter

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

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Equilibrium

Some More Questions From Equilibrium Chapter

Name the factors on which vapour pressure of any liquid depends.

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

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Equilibrium

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass Calculate Kc for this reaction at the above example.

Some More Questions From Equilibrium Chapter

Name the factors on which vapour pressure of any liquid depends.

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation