Chemical and Ionic Equilibrium MCQs Question and Answer in Engineering Chemistry

A. 6.0 × 10^-4, 3.0 × 10^-4, 1.5 × 10^-3

B. 6.0 × 10^-4, 3.0 × 10^-4, 9 × 10^-4

C. 5.0 × 10^-4, 3.5 × 10^-4, 13.5 × 10^-3

D. 5.0 × 10^-4, 3.5 × 10^-4, 8.5 × 10^-4

A. 0.186

B. 0.512

C. 0.0512

D. 0.51/1.86

A. 2SO₃(g) $$ \rightleftharpoons $$ 2SO₂(g) + O₂(g)

B. H₂(g) + l₂(s) $$ \rightleftharpoons $$ 2HI(g)

C. C(s) + H₂O(g) $$ \rightleftharpoons $$ CO(g) + H₂(g)

D. 3Fe(s) + 4H₂O(g) $$ \rightleftharpoons $$ Fe₃O₄(s) + 4H₂(g)

A. 2

B. 3

C. 4

D. 5

A. Gas particles have mass but no volume

B. Particles are in a Brownian motion between collisions

C. During the collision, the system does not lose energy

D. Particles exert same force per unit area on all sides of the container

A. 0

B. 1

C. 2

D. 3

A. $${\left( {\frac{{\partial p}}{{\partial T}}} \right)_V}{\left( {\frac{{\partial T}}{{\partial V}}} \right)_p}{\left( {\frac{{\partial V}}{{\partial p}}} \right)_T} = 0$$

B. $${\left( {\frac{{\partial p}}{{\partial T}}} \right)_V}{\left( {\frac{{\partial T}}{{\partial V}}} \right)_p}{\left( {\frac{{\partial V}}{{\partial p}}} \right)_T} = - 1$$

C. $${\left( {\frac{{\partial p}}{{\partial T}}} \right)_V}{\left( {\frac{{\partial T}}{{\partial V}}} \right)_p}{\left( {\frac{{\partial V}}{{\partial p}}} \right)_T} = + 1$$

D. $${\left( {\frac{{\partial p}}{{\partial T}}} \right)_V}{\left( {\frac{{\partial T}}{{\partial V}}} \right)_p}{\left( {\frac{{\partial V}}{{\partial p}}} \right)_T} = + 2$$

A. > 1

B. < 1

C. 1

D. (1 - b)

A. independent of the molar volume

B. inversely proportional to the molar volume

C. inversely proportional to square of the molar volume

D. directly proportional to the molar volume

A. 16

B. 0.5

C. 2

D. 32