Question

Given the following reaction:
CH3CH2OH(l)+3O2(g)→2CO2(g)+3H2O(l)CH3CH2OH(l)+3O2(g)→2CO2(g)+3H2O(l)

∆H = –2200 kJ/mol CH3CH2OHCH3CH2OH

b) If you increase [CH3CH2OHCH3CH2OH], which of the following will occur? Check all that apply.

Select one or more:

a. the reaction will shift left

b. Heat will be released

c. the reaction will shift right

d. [CO2CO2] will increase

e. [O2O2] will decrease

f. [H2OH2O] will decrease

Answer 1

Answer -

Given,

$\Delta$ H = -2200 kJ/mol

CH3CH2OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l)

According to Le Chatlier Principle, Any Change Caused will be opposed by the reaction by shitfing the Equilibrium.

a)

If CH3CH2OH is increased then the reaction will shift towards Right . So, as to decrease the amount of CH3CH2OH .

So, This OPTION is wrong.

b)

As the equilibrium is shifting in the Right direction i.e. product side. And the reaction is Exothermic. Because heat is a product of this reaction. It will increase.

So, This OPTION is correct.

c)

If CH3CH2OH is increased then the reaction will shift towards Right . So, as to decrease the amount of CH3CH2OH .

So, This OPTION is Correct.

d)

As the equilibrium is shifting in the Right direction i.e. product side. And the reaction is Exothermic. Because CO2 is a product of this reaction. It will increase.

So, This OPTION is correct.

e)

As the equilibrium is shifting in the Right direction i.e. product side. And the reaction is Exothermic. Because O2 is a reactant of this reaction. It will decrease.

So, This OPTION is correct.

f)

As the equilibrium is shifting in the Right direction i.e. product side. And the reaction is Exothermic. Because H2O is a product of this reaction. It will increase.

So, This OPTION is wrong.