1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
morpeh [17]
3 years ago
6

Imagine the reaction A + B ⇌ C + D is at equilibrium, where the forward and reverse reactions are at equal rates. What would hap

pen to the rate of the forward reaction if the [A] were suddenly to increase?
Chemistry
1 answer:
kondaur [170]3 years ago
3 0

Answer:

Rate of forward reaction will increase.

Explanation:

Effect of change in reaction condition on equilibrium is explained by Le Chatelier's principle. According to this principle,

If an equilibrium condition of a dynamic reversible reaction is disturbed by changing concentration, temperature, pressure, volume, etc,  then reaction will move will in a direction which counteract the change.

In the given reaction,

A + B ⇌ C + D

If concentration of A is increase, then reaction will move in a direction which decreases the concentration of A to reestablish the equilibrium.

As concentration A decreases in forward direction, therefore, rate of forward reaction will increase.

You might be interested in
An ideal gas originally at 0.85 atm and 66°C was allowed to expand until its final volume, pressure, and temperature were 94 mL,
svet-max [94.6K]

Answer : The initial volume was, 71.2 mL

Explanation :

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=0.85atm\\V_1=?\\T_1=66^oC=[66+273]K=339K\\P_2=0.60atm\\V_2=94mL\\T_2=43^oC=[43+273]K=316K

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

\frac{0.85atm\times V_1}{339K}=\frac{0.60atm\times 94mL}{316K}

V_1=71.2mL

Therefore, the initial volume was, 71.2 mL

4 0
3 years ago
2) Find the volume that 2.0 moles of H2 will occupy at STP?
Rom4ik [11]

Answer:

V = 44.85 L

Explanation:

Given data:

Volume of H₂ = ?

Number of moles of H₂ = 2.0 mol

Given temperature = 273.15 K

Given pressure = 1 atm

Solution:

Formula:

PV = nRT

P = Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

By putting values,

1 atm × V = 2.0 mol × 0.0821 atm.L/ mol.K  × 273.15 K

V = 44.85 atm.L / 1 atm

V = 44.85 L

4 0
3 years ago
For the reaction ? Fe+? H2o ⇀↽? Fe3o4+? H2 , a maximum of how many grams of fe3o4 could be formed from 354 g of fe and 839 g of
Evgesh-ka [11]

The given reaction is:

3Fe + 4H2O → Fe3O4 + 4H2

Given:

Mass of Fe = 354 g

Mass of H2O = 839 g

Calculation:

Step 1 : Find the limiting reagent

Molar mass of Fe = 56 g/mol

Molar mass of H2O = 18 g/mol

# moles of Fe = mass of Fe/molar mass Fe  = 354/56 = 6.321 moles

# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles

Since moles of Fe is less than H2O;  Fe is the limiting reagent.

Step 2: Calculate moles of Fe3O4 formed

As per reaction stoichiometry:

3 moles of Fe form 1 mole of Fe3O4

Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4

Step 4: calculate the mass of Fe3O4 formed

Molar mass of Fe3O4 = 232 g/mol

# moles = 2.107 moles

Mass of Fe3O4 = moles * molar mass

= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)

 


7 0
3 years ago
Read 2 more answers
10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is
katovenus [111]

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

q=mC(T_2-T_1)

where,

q = heat released = 1.0\times 10^3J=1000J

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

T_2 = final temperature = ?

T_1 = initial temperature = 20°C

Putting values in above equation, we get:

1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC

Hence, the final temperature of the coffee is 43.9°C

6 0
3 years ago
If 200 grams of water is to be heated from 24.0° C to 100.0° C to make a cup of tea, how much heat must be added?
GalinKa [24]
C -------------------
3 0
3 years ago
Read 2 more answers
Other questions:
  • Sand dollars typically live in the intertidal zone. Which adaptation do sand dollars most likely have?
    13·2 answers
  • A tree was chopped down and cut up in a local forest. The tree changed in shape.
    6·2 answers
  • 45.0 mL of a Naoh solution is titrated with 0.12 M HCl. If 29.6 mL is required for neutralization,
    12·1 answer
  • Explain how an ionic compound made up of charged particles can be electrically neutral
    8·2 answers
  • How can looking at the crystal size tell you where an igneous rock formed?
    13·1 answer
  • In the polar molecule HBr, what charge does the H bear
    7·1 answer
  • Why sodium is called active metal​
    9·1 answer
  • How many times more acidic is a solution with a pH of 2 than<br> one with a pH of 4?
    12·1 answer
  • **The diagram below shows the Earth at four different positions in its journey around the Sun.
    5·2 answers
  • Are all molecules of ideal gas in motion with different velocities?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!