CHEMISTRY HIGH SCHOOL

At 700 K the equilibrium constant KC for the reaction between NO(g) and O2(g) forming NO2(g) is 8.7 × 106. The rate constant for the reverse reaction at this temperature is 0.54 M–1s–1. What is the value of the rate constant for the forward reaction at 700 K?

Answers

Answer 1

Answer:

Answer : The value of the rate constant for the forward reaction at 700 K is, $4.70* 10^6$

Explanation :

The given chemical equilibrium reaction is:

$NO(g)+O_2(g)\rightleftharpoons NO_2(g)$

The expression for equilibrium constant is:

$K_c=([NO_2])/([NO][O_2])$

The expression for rate of forward and backward reaction is:

$R_f=K_f[NO][O_2]$

and,

$R_b=K_b[NO_2]$

As we know that at equilibrium rate of forward reaction is equal to rate of backward reaction.

$R_f=R_b$

$K_f[NO][O_2]=K_b[NO_2]$

$(K_f)/(K_b)=([NO_2])/([NO][O_2])$

$(K_f)/(K_b)=K_c$

Given:

$K_c=8.7* 10^6$

$K_b=0.54M^(-1)s^(-1)$

Now put all the given values in the above expression we get:

$(K_f)/(K_b)=K_c$

$(K_f)/(0.54)=8.7* 10^6$

$K_f=4.70* 10^6$

Therefore, the value of the rate constant for the forward reaction at 700 K is, $4.70* 10^6$

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If an excess of sulfric acid reacts with 23.0 grams of sodium chloride how many grams of hydrochloric acid are produced? 2H2SO4 + 4NaCI -> 2H2SO4 + 4HCI

Answers

NaCl = 58.44 g/mol
H₂SO₄ = 98.0 g/mol

H₂SO₄ + 2 NaCl = 2 HCl + Na₂SO₄

98.0 g H₂SO₄ -----------> 2 x 58.44 g NaCl
? g H₂SO₄ ------------> 23.0 g NaCl

Mass of H₂SO₄ = 23.0 x 98.0 / 2 x 58.44

Mass of H₂SO₄ = 2254 / 116.88

=> 19.284 g of H₂SO₄

hope this helps!

Nonmetals make up the largest grouping of elements on the periodic table. answer True or False

Answers

false, nonmetals do not make up the largest grouping of elements on the periodic table

The answer is false metals make up the largest group

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answers

The partial pressure exerted by a gas in a mixture, depends on the mole

fraction of the gas and the pressure exerted by the mixture.

The partial pressure of H₂O is 20 atm.

Reasons:

Given parameters are;

Explosion equation is 4C₃H₅N₃O₉ → 12CO₂(g) + O₂(g) + 6N₂(g) + 10H₂O(g)

Amount of nitroglycerine = 227 g

Molar mass of nitroglycerine = 227 g/mol

Required:

Partial pressure of the water vapor

Solution:

Number of moles of nitroglycerine in the reaction = 1 mole

Therefore;

Number of moles of CO₂ = 12/4 = 3 moles

Number of moles of O₂ = 0.25 moles

Number of nitrogen, N = 1.5 moles

Number of moles of H₂O = 2.5 moles

$Mole \ fraction \ of \ H_2O, \ X_(H_2O) = (2.5)/(2.5 + 1.5 + 0.25 + 3) = (10)/(29)$

According to Raoults law, we have;

The partial pressure of H₂O = $X_(H_2O) * P_$

Therefore, partial pressure of H₂O = $(10)/(29) * 58$ = 20 atm.

Learn more here:

brainly.com/question/10165688

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

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=(227g)/(227g/mol)=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(12)/(4)=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(1)/(4)=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(6)/(4)=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(10)/(4)=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=(2.5)/(2.5+3+0.25+1.5)=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_(H_2O)=X_(H_2O)* p_T$

where,

$p_(H_2O)$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_(H_2O)$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_(H_2O)=X_(H_2O)* p_T$

$p_(H_2O)=0.345* 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

What is the atomic number, icon, mass of potassium?

This should be EASY!

Answers

Answer:

19, K, 39.098

Explanation:

Potassium is a chemical element with symbol K and atomic number 19.

Atomic Mass: 39.098u

What is 19 degrees Celsius in Fahrenheit?

Answers

ºF = ºC x 1.8 + 32

ºF = 19 x 1.8 + 32

ºF = 34.2 + 32

= 66.2ºF

hope this helps!

In a chemical reaction, substance R reacts with compound XY to produce 47.0 g of compound RX. If the theoretical yield of RX is 56.0 g, what is its percent yield?

Answers

The percent yield of a chemical reaction is calculated by dividing the actual yield (determined through experimentation) by the theoretical yield (calculations). The percent yield is a great way of determining the efficiency of a reaction. For this problem, the percent yield is 83.93% as given by the solution:

% yield = (47/56)*100 = 83.93%

Answer:

C. 83.9%

Explanation:

I got it right on edge

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