Consider the following reversible reaction. Upper C (s) + Upper O Subscript 2 Baseline (g) double arrow Upper C Upper O Subscript 2 Baseline (g)

What is the equilibrium constant expression for the given system?

k Subscript e q Baseline = StartFraction left-bracket Upper C Upper O Subscript 2 Baseline right-bracket Over left-bracket Upper C right-bracket left-bracket Upper O Subscript 2 Baseline right-bracket EndFraction
k Subscript e q Baseline = StartFraction left-bracket Upper C Upper O Subscript 2 Baseline right-bracket Over left-bracket Upper O Subscript 2 Baseline right-bracket EndFraction
k Subscript e q Baseline = StartFraction left-bracket Upper C right-bracket left-bracket Upper O Subscript 2 Baseline right-bracket Over left-bracket Upper C Upper O Subscript 2 Baseline right-bracket EndFraction
k Subscript e q Baseline = StartFraction left-bracket Upper O Subscript 2 Baseline right-bracket Over left-bracket Upper C Upper O Subscript 2 Baseline right-bracket EndFraction

Keq = [CO₂]/[O₂]

Explanation:

Step 1: Write the balanced equation for the reaction at equilibrium

C(s) + O₂(g) ⇄ CO₂(g)

Step 2: Write the expression for the equilibrium constant (Keq)

The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:

Keq = [CO₂]/[O₂]

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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