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3 years ago

15

Explain the relationship between forward and reverse

1 answer:

serious [3.7K]3 years ago

8 0

If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position.

For example, if the system is changed in a way that increases the concentration of one of the reacting species, it must favor the reaction in which that species is consumed. In other words, if there is an increase in products, the reaction quotient, Qc, is increased, making it greater than the equilibrium constant, Kc.

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Measure of the average kinetic energy of molecules in an in an object

algol13

Yes because molecules is solid

8 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

3 years ago

Chemical equation to show cdso4 dissolves in water

olga_2 [115]

no se bb si quieres salimos tu sabes

5 0

3 years ago

What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?

cupoosta [38]

Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of $N_2$ = 2.00 M

Concentration of $H_2$ = 2.00 M

Concentration of $NH_3$ = 1.00 M

Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.

The balanced equilibrium reaction is,

$N_2+3H_2\rightleftharpoons 2NH_3$

The expression of reaction quotient for this reaction is,

$Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}$

Now put all the given values in this expression, we get

$Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625$

Therefore, the value of reaction quotient, Q is 0.0625.

3 0

3 years ago

How do you calculate mass using density and volume?

vazorg [7]

D = M/v
M = Dv
V = M/D

7 0

3 years ago