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Chemical Equation MCQ - 1 (Advanced)


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10 Questions MCQ Test Chemistry for JEE Advanced | Chemical Equation MCQ - 1 (Advanced)

Chemical Equation MCQ - 1 (Advanced) for JEE 2022 is part of Chemistry for JEE Advanced preparation. The Chemical Equation MCQ - 1 (Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Chemical Equation MCQ - 1 (Advanced) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chemical Equation MCQ - 1 (Advanced) below.
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Chemical Equation MCQ - 1 (Advanced) - Question 1

X,Y and Z react in the 1 : 1 : 1 stoichiometric ratio. The concentration of X, Y  and Z where found to vary with time as shown in the figure below

Q.

Which of the following equilibrium reaction represents the correct variation of concentration with time.               

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 1

Clearly concentration of Y is not changing with the time hence it will be pure solid or liquid. Concentration of X decreasing hence it will be in reactant and Z will be in product.

Chemical Equation MCQ - 1 (Advanced) - Question 2

X,Y and Z react in the 1 : 1 : 1 stoichiometric ratio. The concentration of X, Y  and Z where found to vary with time as shown in the figure below

Q.

Value of the equilibrium constant (Kc) for the equilibrium represented in above sketch will be.

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 2

Chemical Equation MCQ - 1 (Advanced) - Question 3

X,Y and Z react in the 1 : 1 : 1 stoichiometric ratio. The concentration of X, Y  and Z where found to vary with time as shown in the figure below

Q.

If above equilibrium is established in a 2L container by taking reactants in sufficient amount then how many moles of component Y must have reacted to establish the equilibrium.

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 3

From the graph we can see that 4 mole/litre of Z is formed therefore      4 × 2 moles of solid will be required.

Chemical Equation MCQ - 1 (Advanced) - Question 4

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx            

Total no. of moles  

Observed molecular weight or molar mass of the mixture 

 

Q.

A sample of mixture of A(g), B(g) and C(g) under equilibrium has a mean molecular weight (observed) is 80.

The equilibrium is 

Find the degree of dissociation α for A(g).

                   

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 4

Chemical Equation MCQ - 1 (Advanced) - Question 5

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx            

Total no. of moles  

Observed molecular weight or molar mass of the mixture 

Q.

If the total mass of the mixture in the above case is 300 gm, the moles at C(g) present are.

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 5

Mols of C = a α

 

Chemical Equation MCQ - 1 (Advanced) - Question 6

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx            

Total no. of moles  

Observed molecular weight or molar mass of the mixture 

Q.

The K for the reaction    is 640 mm at 775 K. The percentage dissociation of N2O4 at equilibrium pressure of 160 mm is :

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 6

Chemical Equation MCQ - 1 (Advanced) - Question 7

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx            

Total no. of moles  

Observed molecular weight or molar mass of the mixture 

 

Q.

x (degree of dissociation) varies with D/d in the above reaction according to :

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 7

Chemical Equation MCQ - 1 (Advanced) - Question 8

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus               pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula. The observed densities decrease towards a limit as the temperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.

Examples :

With increase in the number of molecules, the volume increases (pressure  remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,

t = 0 a

t = teq       a  — x               nx            

Total no. of moles  

Observed molecular weight or molar mass of the mixture 

 

Q.

The equation    is correctly matched for :

 

Detailed Solution for Chemical Equation MCQ - 1 (Advanced) - Question 8

Hence one mole of reactant should produce total n moles of product.

Chemical Equation MCQ - 1 (Advanced) - Question 9

Match the following

Chemical Equation MCQ - 1 (Advanced) - Question 10

Match the following (multiple)

                Left column : Represents an equilibrium situratton through a chemical equation and below each equation a stimulus is given which may or may not

                disturb the equilibrium situration.

                Right coloumn : Represents the responses immediately after the    disturbance is created.

                 With                                                                      R1 : Rate of forward reaction.

                                                                        Rb Rate of backward reaction.

                                                                        Q : Reaction quotient

                                                                        K : Equilibrium constant

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