The given sentence is part of a longer question.
I found this question with the same sentence. So, I will help you using this question:
For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>). If </span>Kp = 0.15, which statement is true of the reaction mixture before
any reaction occurs?
(a) Q = K<span>; The reaction </span>is at equilibrium.
(b) Q < K<span>;
The reaction </span>will proceed to
the right.
(c) Q > K<span>; The reaction </span>will proceed to the left.
The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left,
since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:
Kp is the equilibrium constant in term of the partial pressures of the gases.
Q is the reaction quotient. It is a measure of the progress of a chemical reaction.
The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.
At equilibrium both Kp and Q are equal. Q = Kp
If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,
If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.
Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.
Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
Answer: The concentrations of
at equilibrium is 0.023 M
Explanation:
Moles of
= ![\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bgiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D%5Cfrac%7B10g%7D%7B71g%2Fmol%7D%3D0.14mol)
Volume of solution = 1 L
Initial concentration of
= ![\frac{0.14mol}{1L}=0.14M](https://tex.z-dn.net/?f=%5Cfrac%7B0.14mol%7D%7B1L%7D%3D0.14M)
The given balanced equilibrium reaction is,
![COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)](https://tex.z-dn.net/?f=COCl_2%28g%29%5Crightleftharpoons%20CO%28g%29%2BCl_2%28g%29)
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
![4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}](https://tex.z-dn.net/?f=4.63%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7Bx%29%5E2%7D%7B%280.14-x%29%7D)
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of
at equilibrium is 0.023 M
A catalyst speeds up the rate of reaction so the answer is B.
The ph before the addition of any Koh is<u> 10.105.</u>
Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.
After the addition of 50 ml KOH,
moles of KOH = 50 * 0.13 =<u> 6.5 mmol </u>
<u>moles </u><u>of HClO = 50 * 0.13 = 6.5 mmol </u>
occurred hydrolysis solution,
pH = 0.5(14 + pKa + log [base conjugate])
pH = 0.5(14 + (- log (4 * 10^-8)) + log (6.5/(50 + 50)))
pH = <u>10.105</u>
The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.
The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.
Learn more about concentration here:-brainly.com/question/26255204
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