Equilibrium and Thermodynamics (Chem) Flashcards

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1
Q

How can you tell that a reaction favors the products from a Keq equation?

A

The numerator is greater than the demonimator,thus it will have a large Keq

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2
Q

How can you tell that a reaction favors the reactants from a Keq equation?

A

The numerator is smaller than the demonimator,thus it will have a small Keq

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3
Q

True or False

You can substitute any concentration into a Keq expression and expect to obtain an equilibrium constant

A

False

The concentrations used MUST be equilibrium concentrations

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4
Q

Using the equilibrium equation Keq = kforward / kreverse , write the Keq expression for the reverse reaction of a balanced equation of 2X(g) + 4Y(g) ⇌ 3Z (g)

A

Keq = X2 [Y]4 / [Z]3

where the exponents are derived from the coefficients from the balanced equation

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5
Q

Using the equilibrium equation Keq = kforward / kreverse , write the Keq expression for the of a balanced equation of 2X(g) + 4Y(g) ⇌ 3Z (g)

A

Keq = [Z]3 / X2 [Y]4

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6
Q

Write the general forumula of the Keq of the reverse reaction

A

1/Keq

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7
Q

True or False

Keq can be reported in negative values

A

False

since concentrations are given in units of moles/liter or molarity, it cannot be negative

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8
Q

In the chemical equation SnO2 (s) + 2CO (g) ⇋ Sn (s) + 2 CO2 (g) , what molecules would we exclude from the equilibrium expression and why?

A

SnO2 and Sn, because they are solids (s)

We do not include pure solids or liquids in equilibrium expressions

link KA equilibrium page

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9
Q

When do we use the reaction quotient (Q) instead of the equilibrium constant Keq?

A

When we’re uncertain or do not know the equilibrium concentrations

The equation for the reaction quotient is the same as the equation used for Keq, but the concentrations can be from any moment in time, not necessarily at equilibrium. Similarly, the Q expression should not include pure solids, liquids, or solvents

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10
Q

When Q < Keq…

A

The ratio of products to reactants is lower than it would be in equilibrium

The reaction will proceed forward

pg. 52 of chem book for examples

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11
Q

When Q > Keq…

A

The ratio of products to reactants is* greater* than it would be in equilibrium

The reaction will proceed in the reverse

pg. 52 of chem book for examples

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12
Q

When Q = Keq…

A

The reaction is at equilibrium

pg. 52 of chem book for examples

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13
Q

Combustion and neutralization reactions are ____ thermic

A

Exothermic (-ΔH)

releases heat

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14
Q

Given Le Chatlier’s principle, an increase in temperature has the effect of:

A

A shift towards products if ΔH > 0, and a shift towards reactants if ΔH < 0

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15
Q

Given Le Chatlier’s principle, an decrease in temperature has the effect of:

A

A shift towards reactants if ΔH > 0, and a shift towards products if ΔH < 0

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16
Q

True or False

Catalysts increases the rate of both a forward and reverse reactions, causing it to reach equilibrium more quickly

A

True

Note: this is only before the reaction is already at equilibrium!

17
Q

True or False

Catalysts change the relative amounts of products and reactants

A

False

They only change the reaction rates

18
Q

Explain the equation ΔE = Q -W . Which law of thermodynamics does this correspond to?

A

Net change in total energy = Heat added - work done

first law of thermodynamics

if the system is being heated, Q is positive, but if heat is being lost from the system Q is negative. When work is being done by the system on its surroundings, W is positive. This is because work done by te system causes the system’s internal energy to decrease. Work done on the system is negative in the above equation. i.e, when work is doneon a system, its internal energy increases since -(-W) is a positive number.

19
Q

What formula can be used to calculate the necessary quantity of heat to raise the temperature of a material?

A

q = mcΔT

aka the calorimetry equation

Where:

q = heat required (J)
m = mass of substance present (g)
c = substance’s specific heat (J/g*ºC)
ΔT = temperature change (ºC)

20
Q

Write the equation for calculating the changes in enthalpy

A

ΔH = ΔE + PΔV

Where ΔE = change in internal energy of the system

Enthalpy is measured in Joules (J) or kilojoules (kJ)

21
Q

What is considered “standard conditions” for standard enthalpy/heat of formation?

A

1 atm and 298 Kelvins

22
Q

Name the equation

ΔHrxn = ΔHproducts - ΔHreactants

ΔHrxn = ΔH (bonds broken) - ΔH (bonds formed)

A

Standard Enthalpy of formation

23
Q

ΔSrxn = ΔSproducts - ΔSreactants

A

Standrard Entropy

think: second law of thermodynamics

24
Q

Name the formula

quantifies the amount of usable [free] energy (energy that can do work) in a system; the change in free energy of a system as it goes from some initial state, such as all reactants, to some other, final state, such as all products

measures the spontaneity of a rxn

A

Gibbs Free Energy

T = temperature (in kelvin)
S = entropy

| H = enthalpy

## Footnote

units: kJ/mol rxn


25
Q

ΔG°rxn = ΔG°products - ΔG°reactants

A

Gibbs free energy equation for standard conditions

standard formation of Gibbs free energy

26
Q

The ΔG of a reaction at equilibrium is…

A

Zero

because there is no overall change in the system (reactants arent moving to products and vice versa), there is no change in free energy

27
Q

ΔG°rxn = -RTlnKeq

A

Gibbs free energy wrt the equilibrium constant

R = gas constant
T = temp in Kelvins

note: ln is natural log; ln of 1 is 0, ln of a number greater than 1 is positive, and the ln of a number less than 1 is negative

28
Q

ΔGrxn = ΔG°rxn + RTlnQ

A

can be used to assess a rxn under non-standard conditions

29
Q

A reaction in which the product ratio is determined by the relative stability of the products

A

Thermodynamic control

30
Q

At low temperature, the reaction is under ____ (rate, irreversible conditions) and the major product is that from the fastest reaction.

A

Kinetic control

pg. 65

31
Q

the kinetic product forms more quickly/slowly , is less/more thermodynamically favorable, has a lower/higher activation energy, and is favored at low/high temperatures

A

forms more quickly
less thermodynamically stable
lower activation energy
favord at low temperatures

32
Q

the thermodynamic product forms more quickly/slowly , is less/more thermodynamically favorable, has a lower/higher activation energy, and is favored at low/high temperatures

A

forms more slowly
less thermodynamically stable
lower activation energy
favored at low temperatures