Question

In the gas phase at 400 ∘C, isopropyl alcohol (rubbing alcohol) decomposes to acetone, an important industrial solvent: (CH3)2CHOH(g)⇌ (CH3)2CO(g)+H2(g) ΔH∘ = +57.3kJ Isopropyl alcohol Acetone Does the amount of acetone increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?

Multiple choice

Part A

The temperature is increased.

a) The amount of acetone will increase.

b) The amount of acetone will decrease.

c) The amount of acetone will remain the same. The volume is increased.

Part B

The volume is increased.

a) The amount of acetone will increase.

b) The amount of acetone will decrease.

c) The amount of acetone will remain the same.

Part C

Argon is added.

a) The amount of acetone will increase.

b) The amount of acetone will decrease.

c) The amount of acetone will remain the same.

Part D

H2 is added. is added.

a) The amount of acetone will increase.

b) The amount of acetone will decrease.

c) The amount of acetone will remain the same.

Part E

A catalyst is added.

a) The amount of acetone will increase.

b) The amount of acetone will decrease.

c) The amount of acetone will remain the same.

Answer 1

(CH3)2CHOH(g)? (CH3)2CO(g)+H2(g) ?H? = +57.3kJ

?H? is positive that means reaction is endo thermic reaction

Part A

Acetone will increase

because for endo thermic reaction if you increase the temperature reaction will favoured to wards right side

Part B

The amount of acetone will decrease.

if you increase the volume reaction is backward means no of moles will decrease

PartC

The amount of acetone will remain the same.

Argon is a nuetral gas it doesnt affect the equilibrium

PartD

The amount of acetone will decrease

if you add the right side component extra reaction shift backward

Part D

c) The amount of acetone will remain the same.

catalyst will only increase the speed of the reaction it doesnt effect the equilibrium