Question

Co(g)+2h2(g)⇌ch3oh(g) a reaction mixture in a 5.20-l flask at a certain temperature initially contains 26.6 g co and 2.32 g h2. At equilibrium, the flask contains 8.64 gch3oh. Calculate the equilibrium constant (kc)for the reaction at this temperature.

Answer 1

Answer: $k_c$ for the reaction is 27.78

Explanation:

For $CH_3OH$,

We are given equilibrium mass of $CH_3OH$ to be 8.64g.

Moles can be calculated by:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$  .....(1)

Molar mass of $CH_3OH$ = 32.04 g/mol

$\text{Moles of }CH_3OH=\frac{8.64g}{32.04g/mol}=0.2696moles$

Molarity can be calculated by:

$Molarity=\frac{Moles}{\text{Volume (in Liters)}}$     ....(2)

Volume given = 5.20 L

$\text{Molarity of }CH_3OH=\frac{0.2969moles}{5.20L}=0.0518M$

For CO,

Given mass = 26.6 g

Molar mass of CO = 28 g/mol

Using equation 1, we get

$\text{Initial moles of CO}=\frac{26.6g}{28g/mol}=0.95moles$

Using equation 2, we get

$Molarity=\frac{0.95moles}{5.20L}=0.1826M$

For $H_2$,

Given mass = 2.32 g

Molar mass of $H_2$ = 2 g/mol

Using equation 1, we get

$\text{Initial moles of }H_2=\frac{2.32g}{2g/mol}=1.16moles$

Using equation 2, we get

$Molarity=\frac{1.16moles}{5.2L}=0.223M$

For the reaction,

                             $CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

at $t=0$                0.1826M    0.223M          0

at $t=t_{eq}$             (0.1826-x)  (0.223-2x)       x

Here, $x=[CH_3OH]=0.0518M$

$[CO]=0.1826-x=(0.1826-0.0518)M=0.1308M$

$[H_2]=0.223-2x=0.223-2(0.0518)=0.1194M$

Equilibrium Constant, $k_c$ for this given reaction is written as:

$k_c=\frac{[CH_3OH}}{[CO][H_2]^2}$

Putting the concentration or molarity values, we get

$k_c=\frac{(0.0518)}{(0.1308)(0.1194)^2}$

$k_c=27.78$

Answer 2

Answer:

Kc for the reaction is 27.78.

Explanation:

Given: Volume of concentration reaction is 5.20 L and initial concentration of CO is 26.6g and H2 is 2.32 g which are reactants and concentration of product that is CH3OH is 8.64 g.

To find: Kc for the reaction mixture which is at equilibrium at a given temperature and pressure.

Step Wise Solution: For calculating equilibrium constant we first need to calculate concentration of both reactant and products which can be calculated by molarity.

Molarity = moles of substance/ Volume where moles can be calculated as moles = given mass of substance / gram molecular mass or molar mass.

Let us calculate moles of both reactant and product.

Molar Mass of CO = 28 g/mol

Given mass = 26.6 g

So moles of CO = 26.6/28 = 0.95 moles

Volume given is 5.20 L so molarity = moles / volume = 0.95 / 5.20 = 0.1826 M

Now, similarly for H2 Given mass = 2.32 g

Molar mass of H2 = 2 g/mol

Moles of H2 = 2.32/2 = 1.16 moles

Molarity = Moles/Volume = 1.16/5.20 = 0.223 M

For product that is CH3OH we have:-

Molar mass of CH3OH = 32.04 g/mol

Given Mass = 8.64 g So moles = 8.64/32.04 = 0.269 moles

Molarity = moles / volume = 0.269/ 5.20 = 0.0518 M

For the reaction, CO     +      2H2-----------> CH3OH

at t=0(initially)   0.1826M    0.223M          0

at t=t(eq)            (0.1826-x)  (0.223-2x)       x

(Let concentration of product at equilibrium be x)

Here, x=[CH3OH]=0.0518M at equilibrium

So, [CO]=0.1826-x=(0.1826-0.0518)M=0.1308M

Similarly for [H2]=0.223-2x=0.223-2(0.0518)=0.1194M

We know that equilibrium constant

Kc = Concentration of product/ Concentration of reactant

So Kc = [CH3OH]/[CO][H2]^2

Substituting the values we have:

Kc = 0.0518/0.1308(0.1194)^2 so after calculating Kc = 27.78

Note: Since equilibrium constant is a ratio it has no unit.

Code: SPJ2