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3 years ago

Which one of the following will change the value of anequilibrium constant?

Chemistry

1 answer:

IRINA_888 [86]3 years ago

6 0

Answer:

d. changing temperature

Explanation:

The thermodynamic equilibrium constant K is defined as a quantity characterizing the equilibrium of a chemical reaction. For a reaction where concentrations are in equilibrium:

aA + bB ⇄ cC + dD

The equilibrium constant is:

$k = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

Thus, the equilibrium constant will change if:

a. Varying the initial concentration of reactants . FALSE. The k constant doesn't depend of initial concentrations but concentration in equilibrium does.

b. Adding other substances that do not react with any of thespecies involved in the equilibrium . FALSE. The equilibrium constant just depends of substances that are involved in the equilibrium

c. Varying the initial concentration of products . FALSE. Again, equilibrium constant doesn't depend of initial concentrations.

d. Changing temperature . TRUE. As a thermodynamic constant, k depends of temperature thus:

$K = e^(-dG/RT)$

e. Changing the volume of the reaction vessel. FALSE. The changing in the volume of the reaction vessel will change just the initial concentrations of the reactants.

I hope it helps!

You might be interested in

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0

3 years ago

Read 2 more answers

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:

Margarita [4]

Answer:

108.6 g

Explanation:

2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the PV=nRT formula to calculate the number of nitrogen moles:

P = 1.00 atm

V = 56.0 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 0 °C ⇒ 0 + 273.2 = 273.2 K

Inputting the data:

1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K

n = 2.5 mol

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

2.5 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.67 mol NaN₃

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

1.67 mol * 65 g/mol = 108.6 g

6 0

3 years ago

Mrs. DeFord wanted to know whether or not her students would do better on a quiz if she promised them candy.

marta [7]

Answer:

a.) Independent Variable: # of candy bars promised to each group

[ The Independent Varaiable is what you change in the experiment]

b.) Dependent Variable: Quiz Scores

[ The Dependent Variable is what you're testing in the experiment; what the experiment should affect]

c.) Constant(s): Same Quiz, same number of gender kids in each group, same age kids in each group, same ability, and same background. [and same time, I'm assuming.]

[Constants are what you keep the same in the experiment; what you're not changing.]

d.) Testable Question: Will promising kids candy make them do better on tests and quizzes? [or something along this lines of this]

[The Testable Question is what you're trying to find out in the experiment]

e.) Hypothesis: The more candy the students were promised, the better results Mrs. DeFord would get from them.

[The Hypothesis is what the person performing the experiment expects will happen; an educated guess]

f. Formal Conclusion: Data shows that kids who were promised more candy had a better average than kids who were promised less candy/none. Mrs. DeFord's hypothesis was correct, since she assumed the more candy the students were promised, the better they would do on the quiz.

[The Formal Conclusion is what you have learned from the experiment, and wether or not the hypothesis was correct or not.]

I hope this helps! :)

6 0

3 years ago

Read 2 more answers

The element X forms ions of charge +2 and the element Y forms ions of charge -3. The most likely formula for a binary compound b

tester [92]

Answer:

X3Y2

Explanation:

To represent the chemical formula of the binary compound, the valencies of both elements need to be exchanged.

Valency is the combining power of elements, ions or radicals. It is the value that shows how an element tend to combine with other elements.

In this particular question, the valence of the element X is +2, this means when it enters into chemical combination, it goes in with the power of positive 2. Same goes for the element Y, when it goes into chemical combination, it is expected that it goes in with the negative power of 3.

To completely write the chemical formula of the compound formed from the combination of both, we simply exchange the valency as said above. It must be noted that while the less electronegative element carry a positive or less negative charge , the more electronegative element will carry a negative or more negative charge.

We write the less electronegative element before the more electronegative element while writing the formula of the compound

5 0

4 years ago

Read 2 more answers

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the

saveliy_v [14]

Answer:

The answer is "Option b"

Explanation:

In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:

Formula:

$\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter$

$\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\$

$=\ 0.0025 \ moles$

$\ moles \ in \ HF \ = 0.126M \times 0.0250 L$

$= 0.00315 \ moles$

$\ moles \ in \ HCl = 0.0100M \times 0.00500 L$

$= 0.00005 \ moles$

$\ Reaction: \\\\F - + H+ \rightarrow HF$

$\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245$

$\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V$

$=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M$

8 0

3 years ago