Chemistry Chapter 6 Entry Test MCQs

Topic 6

a. Whole reactant does not change into
b. It can proceed in both directions product
c. It is applicable to stoichiometric calculation
d. Chemical equilibrium can establish in it

c. It is applicable to stoichiometric calculation

a. NH4NO3 (s) ⇌ N2O (g) + 2H2O (g)
b. 2CO (g) + O2 (g) ⇌ 2CO2 (g)
c. 2H2O2 (1) ⇌ 2H2O (I) + O2 (g)
d. BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)

b. 2CO (g) + O2 (g) ⇌ 2CO2 (g)

a. The amount of reactant = The amount of product
b. It is macroscopic property
c. Established from either side of reactant or product
d. It is established in closed system

a. The amount of reactant = The amount of product

a. Increase four times
b. Decrease to one half
c. Decrease to one
d. Increase two times

a. Increase four times

a. Temperature
b. Concentration of reactants
c. Concentration of products
d. Pressure

a. Temperature

a. 30
b. 60
c. 120
d. 1 / 60

b. 60

a. PCl5⇌ PCl3 + Cl2
b. H2 + I2 ⇌ 2HI
c. 2SO2 + O2 ⇌ 2SO3
d. N2 + 3H2 ⇌ 2NH3

b. H2 + I2 ⇌ 2HI

a. dm3mol-2
b. mol dm-3
c. dm6mol-2
d. dm3mol

d. dm3mol

a. PCl5 ⇌ PCl3 + Cl2
b. H2 + I2 ⇌ 2HI
c. MgCO3 ⇌ MgO+ CO2
d. N2 + 3H2 ⇌ 2NH3

c. MgCO3 ⇌ MgO+ CO2

a. The enthalpy change is zero
b. The activation energies for both forward and reverse reactions are zero
c. Its rate constants do not vary with temperature
d. There are equal number of moles of reactants and products in it

a. The enthalpy change is zero

a. It is associated with ΔH
b. It may or may not have unit
c. It tells us about rate of reaction
d. It depends on equilibrium concentration

c. It tells us about rate of reaction

a. Decrease of pressure favours backward reaction
b. Increase of pressure favours forward reaction
c. Increase of volume favours backward reaction
d. Decrease of pressure favours forward reaction

d. Decrease of pressure favours forward reaction

a. 2 × 10-2
b. 1 × 10-2
c. 8 × 10-2
d. 4 × 10-2

b. 1 × 10-2

a. 4.8 × 109 M
b. 2.4 × 109 M
c. 2.8 × 109 M
d. 2.8 × 109 M

a. 4.8 × 109 M

a. Kp = Kc(RT)+1
b. Kc = Kp
c. Kp = Kc(P/N)-2
d. Kp = Kc(RT)-1

d. Kp = Kc(RT)-1

a. Decrease by four times
b. Increase nine times
c. Increase four times
d. Be doubled

b. Increase nine times

a. Kc = Kp (RT)-1/2
b. Kp = Kc√𝑅𝑇
c. Kp = Kc RT
d. Kc = Kp√𝑅𝑇

d. Kc = Kp√𝑅𝑇

a. 1/RT
b. RT
c. 1/ (RT)1/2
d. (RT)1/2

c. 1/ (RT)1/2

a. 0.5
b. 2.25
c. 1.5
d. 1

b. 2.25

a. Moderate Kc Value
b. Kc is very large
c. Can’t determined
d. Kc is very small

b. Kc is very large

a. 75
b. 125
c. 100
d. 50

d. 50

a. Concentration
b. Pressure
c. Temperature
d. Catalyst

c. Temperature

a. depend upon equilibrium concentrations
b. May or may not have a units
c. Thermodynamic property
d. All of the above

d. All of the above

a. PCl5(g) ⇌ PCl3(g) + Cl2(g)
b. N2(g) + 3H2(g) ⇌ 2NH3(g)
c. 2SO2(g) + O2(g) ⇌ 2SO3(g)
d. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

b. N2(g) + 3H2(g) ⇌ 2NH3(g)

a. 12.5
b. 25
c. 50
d. 5

b. 25

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