Question

consider the equilibrium system described by the chemical reaction below. if 0.1908 moles of co₂, 0.0908 moles of h₂, 0.0092 moles of co, and 0.0092 moles of h₂o are present in a 2.00 l reaction vessel at equilibrium at 300 k, what is the value of kc of the reaction at this temperature? co₂(g) h₂(g) ⇌ co(g) h₂o(g)

Answer 1

Here,

CO2 = $\frac{moles}{volume}$ = $\frac{0.1908}{2}$ = 0.0954M

H2 = $\frac{0.1908 }{2}$  = 0.0454M

CO = $\frac{0.0092}{2}$ = 0.0046M

H2O = $\frac{0.0092}{2}$ =  0.0046M

Now, Kc = $\frac{ 0.0046 * 0.0046}{0.0954 *0.0454M}$

          Kc = 4.886 ×$10^{-3}$    

What is equilibrium system ?

When the rates of forward and reverse reactions are equal, a system is said to be in equilibrium. The rate of the forward reaction increases when more reactant is added. As the reverse reaction rate remains constant at first, the equilibrium appears to shift toward the product, or right, side of the equation.

Hence 4.886 ×$10^{-3}$     is a correct answer.

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