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3 years ago

Explain the relationship between forward and reverse reactions at equilibrium

Chemistry

1 answer:

faust18 [17]3 years ago

8 0

Answer:

In a chemical reaction, chemical equilibrium is the state in which the forward reaction rate and the reverse reaction rate are equal. The result of this equilibrium is that the concentrations of the reactants and the products do not change.

Explanation:

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How are atomic nuclei developed

mezya [45]

Answer:

An atom is composed of a positively-charged nucleus, with a cloud of negatively-charged electrons surrounding it, bound together by electrostatic force. ... Protons and neutrons are bound together to form a nucleus by the nuclear force.

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4 years ago

A. Electrons have a charge of_____

Dmitry [639]

Electrons: negative
Protons: positive
Neutrons: nuetral

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3 years ago

Read 2 more answers

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

7 0

3 years ago

A compound is 80.0% carbon and 20.0% hydrogen by mass. assume you have a 100.-g sample of this compound. the molar mass of the c

ch4aika [34]

Basis of the calculation: 100g

For Carbon:
Mass of carbon = (100 g)(0.80) = 80 g
Number of moles of carbon = (80 g)(1 mole / 12g) = 20/3

For Hydrogen:
Mass of hydrogen = (100 g)(0.20) = 20 g
Number of moles of hydrogen = (20 g)(1 mole / 1 g) = 20

Translating the answer to the formula of the substance,
C20/3H20

Dividing the answer,
CH3

The molar mass of the empirical formula is:
12 + 3 = 15 g/mol

Since, the molar mass given for the molecular formula is 30.069 g/mol, the molecular equation is,
C2H6

ANSWER: C2H6

4 0

4 years ago

Can u help?? It's science and I'm in 6th grade

kvv77 [185]

Yeah I can help I'm an A student in science

6 0

4 years ago