What is the symbol for voltage? A. p B. V C. I D. v

Consider the reaction: NO2(g) + CO(g) ⇌ NO(g) + CO2(g) Kc = 0.30 at some temperature. If the initial mixture has the concentrati

Stolb23 [73]

This is an incomplete question, here is a complete question.

Consider the reaction: $NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)$

Kc = 0.30 at some temperature.

If the initial mixture has the concentrations below, the system is_______.

Chemicals Concentration (mol/L)

- NO₂ 0.024

- CO 0.360

- NO 0.180

- CO₂ 0.120

Possible answers:

1) not at equilibrium and will remain in an unequilibrated state.

2) not at equilibrium and will shift to the left to achieve an equilibrium state.

3) not at equilibrium and will shift to the right to achieve an equilibrium state.

4) at equilibrium

Answer : The correct option is, (2) not at equilibrium and will shift to the left to achieve an equilibrium state.

Explanation:

Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

First we have to determine the value of reaction quotient (Qc).

The given balanced chemical reaction is,

$NO_2(g)+CO(g)\rightleftharpoons NO(g)+CO_2(g)$

The expression for reaction quotient will be :

$Q_c=\frac{[NO][CO_2]}{[NO_2][CO]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q_c=\frac{(0.180)\times (0.120)}{(0.024)\times (0.360)}=2.5$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K$ that means product = reactant. So, the reaction is in equilibrium.

The given equilibrium constant value is, $K_c=0.30$

From the above we conclude that, the $Q>K$ that means reactant < product. So, the reaction is reactant favored that means reaction must shift to the reactant or left to be in equilibrium.

Hence, the correct option is, (2) not at equilibrium and will shift to the left to achieve an equilibrium state.

8 0

3 years ago

Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop

jek_recluse [69]

(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is 1.3 x $10^{-5}$ .

The initial concentration of propanoic acid is .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid. It ionizes partially in water as follows:

The expression for acid dissociation constant is,

…… (1)

Here,

is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

is the equilibrium concentration of propanoic acid.

ICE table (1):

Refer ICE table (1),

Substitute the values form the ICE table (1) in equation (1).

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

Rearrange above equation for x.

…… (2)

Substitute for in equation (2) to calculate the value of x.

Therefore, from the ICE table (1) the concentration of hydronium ion is,

.

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

…… (3)

Substitute for in equation (3) to calculate the pH of the solution.

(b)

Given information:

The value of acid ionization constant for propanoic acid is .

The initial concentration of sodium propanoate is .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

…… (4)

Propanoic acid is a weak acid. It ionizes partially in water as follows:

…… (5)

Dissociation reaction for water is written as follows:

…… (6)

From equation (4), (5), and (6) the relationship between and is,

…… (7)

Substitute for and for in equation (7).

ICE table (2):

The expression for base dissociation constant is,

…… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

is the equilibrium concentration of propanoic acid.

From the ICE table (2),

Substitute the values form the ICE table (2) in equation (8).

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

Rearrange above equation for y.

…… (9)

Substitute for in equation (9) to calculate the value of y.

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

…… (10)

Substitute for in equation (10) to calculate pOH of the solution.

The relation between pH and pOH is as follows:

…… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

(c)

Given information:

The value of acid ionization constant for propanoic acid is .

The initial concentration of sodium propanoate is .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

…… (12)

The negative logarithm of acid ionization constant is equal to .

…… (13)

Substitute for in equation (13).

Substitute for , for and 4.9 for in equation (12).

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

3 0

4 years ago

Read 2 more answers

Which of the examples is potassium? es ) A) B) B C)

Nikitich [7]

Answer:

examples of things which contain potassium are:

green vegetables

root vegetables

fruits

potassium chloride

potassium sulphate

Explanation:

if you need a specific answer please send the options

7 0

4 years ago

Jamie rolls a 6-sided die 30 times and determines that the experimental probability of rolling a 2 is Start Fraction 1 over 15 E

anyanavicka [17]

Answer:

Conduct more trials

Explanation:

Theoretical Probability can be defined as what someone is expecting to happen

Experimental Probability on the other hand, is defined as what actually happens.

Probability is usually calculated in the same way for experimental probability and that of theoretical probability. You divide the total number of possible ways in which a particular outcome can happen, by the total number of outcomes itself.

In Experimental probability, the more times a probability is tried, it gets closer and even more closer to theoretical probability.

So, for the question, Jamie should improve the number of tries more, so as to get his experimental probability results to be closer to the theoretical probability result.

8 0

3 years ago

What is the difference between an orbit in the Bohr model of the hydrogen atom and an orbital in the quantum mechanical model

adelina 88 [10]

Answer:

See explanation

Explanation:

In Bohr's theory, electrons are found in specific regions in space called orbits. These orbits are also called energy levels. An electron may move from one energy level to another by absorbing or emitting energy.

In the wave mechanical model, electrons are not found in a particular region in space according to Heisenberg's uncertainty principle.

We rather define a certain region in space where there is a high probability of locating the electron. This region in space where there is a high probability of locating the electron is called an orbital.

Hence, in the Bohr's model of the atom,electrons can surely be found in orbits while in the wave mechanical model, the orbital is a probability function that describes a region in space where an electron may be found.

4 0

3 years ago