Co(g)+2h2(g)ch3oh(g)
Answers
The rate of forward reaction will become 9 times.
Explanation: Rate law says that rate of a reaction is directly proportional to the active mass of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
CO(g)+2H_2\rightarrow CH_3OH
Rate=r=k[CO]^1[H_2]^2
k= rate constant
Given: Active mass of CO is kept constant and active mass of H_2 is tripled
{\text {new rate}}=r'=k[CO]^1[3\times H_2]^2
r'=k[CO]^1[3]^2\times [H_2]^2
r'=k[CO]^19\times [H_2]^2
r'=9\times r
Thus new rate will become 9 times the original rate.
Answer: k_ck
c
for the reaction is 27.78
Explanation:
For CH_3OHCH
3
OH ,
We are given equilibrium mass of CH_3OHCH
3
OH to be 8.64g.
Moles can be calculated by:
Moles=\frac{\text{Given mass}}{\text{Molar mass}}Moles=
Molar mass
Given mass
.....(1)
Molar mass of CH_3OHCH
3
OH = 32.04 g/mol
\text{Moles of }CH_3OH=\frac{8.64g}{32.04g/mol}=0.2696molesMoles of CH
3
OH=
32.04g/mol
8.64g
=0.2696moles
Molarity can be calculated by:
Molarity=\frac{Moles}{\text{Volume (in Liters)}}Molarity=
Volume (in Liters)
Moles
....(2)
Volume given = 5.20 L
\text{Molarity of }CH_3OH=\frac{0.2969moles}{5.20L}=0.0518MMolarity of CH
3
OH=
5.20L
0.2969moles
=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}=0.95molesInitial moles of CO=
28g/mol
26.6g
=0.95moles
Using equation 2, we get
Molarity=\frac{0.95moles}{5.20L}=0.1826MMolarity=
5.20L
0.95moles
=0.1826M
For H_2H
2
,
Given mass = 2.32 g
Molar mass of H_2H
2
= 2 g/mol
Using equation 1, we get
\text{Initial moles of }H_2=\frac{2.32g}{2g/mol}=1.16molesInitial moles of H
2
=
2g/mol
2.32g
=1.16moles
Using equation 2, we get
Molarity=\frac{1.16moles}{5.2L}=0.223MMolarity=
5.2L
1.16moles
=0.223M
For the reaction,
CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)CO(g)+2H
2
(g)⇌CH
3
OH(g)
at t=0t=0 0.1826M 0.223M 0
at t=t_{eq}t=t
eq
(0.1826-x) (0.223-2x) x
Here, x=[CH_3OH]=0.0518Mx=[CH
3
OH]=0.0518M
[CO]=0.1826-x=(0.1826-0.0518)M=0.1308M[CO]=0.1826−x=(0.1826−0.0518)M=0.1308M
[H_2]=0.223-2x=0.223-2(0.0518)=0.1194M[H
2
]=0.223−2x=0.223−2(0.0518)=0.1194M
Equilibrium Constant, k_ck
c
for this given reaction is written as:
Putting the concentration or molarity values, we get
k_c=\frac{(0.0518)}{(0.1308)(0.1194)^2}k
c
=
(0.1308)(0.1194)
2
(0.0518)
k_c=27.78k
c
=27.78