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kenny6666 [7]
3 years ago
12

Chemistry problem H=WQx solve for x

Chemistry
2 answers:
Lana71 [14]3 years ago
8 0
You just have to isolate x, so just divide by WQ on both sides to get:
x =  \frac{H}{WQ}
DaniilM [7]3 years ago
3 0

Answer:

\x=\frac{\text{H}}{\text{WQ}}

Explanation:

The given equation is \text{H}=\text{WQ}x

We need to solve the given equation for x.

In order to solve for x, we need to transpose WQ from RHS to LHS.

So, \text{H}=\text{WQ}x

Divide both sides of the equation by WQ we get

\\frac{\text{H}}{\text{WQ}}=\frac{\text{WQ}}{\text{WQ}}x

So, \x=\frac{\text{H}}{\text{WQ}}

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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
Troyanec [42]

[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3} at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

  • [\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3};
  • [\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3};
  • [\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}.

There was no Y to start with; its concentration could only have increased. Let the change in [\text{Y}] be +x \; \text{mol}\cdot \text{dm}^{-3}.

Make a \textbf{RICE} table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in [\text{X}] will be +2\;x \; \text{mol}\cdot \text{dm}^{-3} and the <em>change</em> in [\text{Z}] will be -2\;x \; \text{mol}\cdot \text{dm}^{-3}.

\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}.

Add the value in the C row to the I row:

\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 + 2\;x & &x&&3-2\;x\end{array}.

What's the equation of K_c 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}]}.

K_c = 2.25. 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.

(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0.

The degree of this polynomial is three. Plot the equation y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9 on a graph and look for any zeros. There's only one zero at x \approx 0.337. All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: 0.337\;\text{mol}\cdot\text{dm}^{-3}.

7 0
3 years ago
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