Chemistry, 05.02.2020 10:46, tintinayeir2567
The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2 mole of x and 3 mole of z in a one litre vessel?
Answers: 3
![[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}](/tpl/images/0503/5677/4561d.png)
at equilibrium.![[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}](/tpl/images/0503/5677/16f3a.png)
;![[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}](/tpl/images/0503/5677/36c99.png)
;![[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}](/tpl/images/0503/5677/168fc.png)
.![[\text{Y}]](/tpl/images/0503/5677/40b69.png)
be 
. 
table.![[\text{X}]](/tpl/images/0503/5677/1967d.png)
will be 
and the change in ![[\text{Z}]](/tpl/images/0503/5677/21145.png)
will be 
.
.
.
for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator. ![K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}](/tpl/images/0503/5677/260f1.png)
.
. As a result,![\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25](/tpl/images/0503/5677/bc2d1.png)
.
.
on a graph and look for any zeros. There's only one zero at 
. All three concentrations end up greater than zero. 
.
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The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration...
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