CHEM1101 Midterm: acids and bases intro to equilibrium

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30 May 2018

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Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. In

the figure above, equilibrium is finally reached in the third picture.

• As a system approaches equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.

• Once equilibrium is achieved, the amount of each reactant and product remains constant.

Writing the Equation for an Equilibrium Reaction

Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we

write its equation with a double arrow: N2O4(g) ⇌2 NO2(g)

Comparing Rates

• Fo the foad eatio NO4g →  NOg • The ate la is Rate = kf[NO4]

• Fo the eese eatio  NOg → NO4g • The ate la is Rate = k[NO]

Therefore, at equilibrium Ratef = Rater kf[N2O4] = kr[NO2]2

• Reitig this, it eoes the epessio fo the euiliiu ostat, Ke.

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Document Summary

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow: n2o4(g) 2 no2(g) Comparing rates: fo(cid:396) the fo(cid:396)(cid:449)a(cid:396)d (cid:396)ea(cid:272)tio(cid:374) n(cid:1006)o4(cid:894)g(cid:895) (cid:1006) no(cid:1006)(cid:894)g(cid:895) the (cid:396)ate la(cid:449) is rate = kf[n(cid:1006)o4, fo(cid:396) the (cid:396)e(cid:448)e(cid:396)se (cid:396)ea(cid:272)tio(cid:374) (cid:1006) no(cid:1006)(cid:894)g(cid:895) n(cid:1006)o4(cid:894)g(cid:895) the (cid:396)ate la(cid:449) is rate = k(cid:396)[no(cid:1006)](cid:1006) Therefore, at equilibrium ratef = rater kf[n2o4] = kr[no2]2: re(cid:449)(cid:396)iti(cid:374)g this, it (cid:271)e(cid:272)o(cid:373)es the e(cid:454)p(cid:396)essio(cid:374) fo(cid:396) the e(cid:395)uili(cid:271)(cid:396)iu(cid:373) (cid:272)o(cid:374)sta(cid:374)t, ke(cid:395). Another equilibrium the haber process: co(cid:374)sider the haber process, which is the industrial preparation of ammonia: n2(g) + 3 h2(g) 2. Nh3(g: the equilibrium constant depends on stoichiometry: The equilibrium constant: co(cid:374)side(cid:396) the ge(cid:374)e(cid:396)alised (cid:396)ea(cid:272)tio(cid:374): a a + b b d d + e e, the equilibrium expression for this reaction would be.

Related Questions

1. Which quantities are equal in a dynamic equilibrium?
a. the rate constants of the forward and reverse reactions
b. the rates of the forward and reverse reactions
c. the concentrations of the reactants and products
d. the volumes of the reactants and products

2. If the rate constant for the forward reaction in N2O4(g)(colorless) ? 2NO2(g) (brown) is larger than the rate constant forthe reverse reaction, will the constant in [NO2]^2 / [N2O4] =kf/kr
be greater than 1 or smaller than 1?

whiteyak435

Learning Goal:

To understand the relationship between the equilibrium constant and rate constants.

For a general chemical equation

A+B?C+D

the equilibrium constant can be expressed as a ratio of the concentrations:

Kc=[C][D][A][B]

If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows:

forward ratereverse rate==kf[A][B]kr[C][D]

where kf and kr are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal:

kf[A][B]=kr[C][D]

Thus, the rate constants are related to the equilibrium constant in the following manner:

Kc=kfkr=[C][D][A][B]

Part A

For a certain reaction, Kc = 7.37

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium , where kf, and kr, are the rate constants for the forward and reverse reactions.