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3 years ago

Which statements are true about reversible reactions?

Chemistry

2 answers:

Alex73 [517]3 years ago

6 0

Answer:

Option D is correct. Reversible reactions exhibit the same reaction rate for forward and reverse reactions at equilibrium.

Explanation:

Reversible reactions can be illustrated with the equation below:

A + B ⇄ C+ D

The reactants, A and B reacts to give the products, C and D. The products then react to give back the reactants.

Reversible reactions can reach equilibrium without a catalyst. Catalyst only speed up the rate at which the equilibrium is reached. This makes option A wrong.

In reversible reaction, both the forward and backward rate are equal. This makes option B wrong.

Option C is also wrong because no side (reactant of product) is favoured in a reversible reaction, unless the system is disturbed.

Option D is correct. Reversible reactions exhibit the same reaction rate for forward and reverse reactions at equilibrium.

option E is wrong.

umka21 [38]3 years ago

5 0

Answer:

Reversible reactions exhibit the same reaction rate for forward and reverse reactions at equilibrium.

Reversible reactions exhibit constant concentrations of reactants and products at equilibrium

Explanation:

A reversible reaction is a reaction that can proceed in both forward and backward direction.

Equilibrium is attained in a chemical system when there is no observable change in the properties of the system.

At equilibrium, a reversible reaction is occurring in at same rate. That is, the forward and backward reaction is occurring at the same rate. As the rate of the forward and backward reaction remains the same, the concentrations of the reactants and products will also be the same in order for the equilibrium to be maintained.

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Select all that apply. Catalysts can save money by essentially lowering the:

k0ka [10]

Answer: Options (a) and (d) are the correct answer.

Explanation:

A catalyst is the substance which helps in increasing the rate of reaction.

Activation energy is the minimum amount of energy required by reactants to start the reaction. On addition of catalyst, the path of reaction changes because the energy barrier gap reduces and hence, the activation energy also decreases.

In the absence of catalyst, we need to increase the temperature so that reaction can occur quickly.

Whereas on addition of catalyst, there is no need to increase the temperature as the catalyst itself is sufficient to increase the rate of reaction. As a result, temperature should be lowered when there is addition of catalyst in the reaction.

Thus, we can conclude that catalysts can save money by essentially lowering the activation energy and temperature required.

4 0

3 years ago

Read 2 more answers

Question - Complete and balance the following chemical equations:

Mekhanik [1.2K]

Answer is because
Please give feedback

6 0

2 years ago

Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.

Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

$\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%$

$\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%$

$\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%$

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0

3 years ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

<u>Answer:</u> The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

3 years ago

Which is an example of a chemical reaction

valkas [14]

Answer:iron and oxygen combining to make rust. vinegar and baking soda combining to make sodium acetate, carbon dioxide and water

Explanation:

7 0

3 years ago