U2-2 - Reaction Feasibility Flashcards

1
Q

Standard values (states, enthalpy, etc) are taken at a temperature of …

A

298 K (25 °C)

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2
Q

Standard state

A

The most stable state at a pressure of 1 atm at 298 K.

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3
Q

Standard enthalpy of formation

A

The enthalpy change when 1 mol of a substance is formed from its elements in their standard states.

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4
Q

For pure elements in their standard states, ∆H°f = …

A

0

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5
Q

Entropy

A

A measure of the degree of disorder of a system

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6
Q

Symbol for entropy

A

S

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7
Q

Which has the highest entropy:

solids, liquids or gases?

A

Gases

(particles free to move = high disorder)

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8
Q

Which has the lowest entropy:

solids, liquids or gases?

A

Solids

(particles can only vibrate, not move freely = low disorder)

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9
Q

What happens to entropy as temperature increases?

A

Entropy increases (more kinetic energy = more disorder)

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10
Q

State the 2nd law of thermodynamics

A

The total entropy of a reaction system and its surroundings always increases for a spontaneous process.

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11
Q

What is a spontaneous process?

A

A process which will happen without having to add heat.

The products are favoured at the conditions under which the reaction is occurring.

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12
Q

State the 3rd law of thermodynamics

A

At 0 K the entropy of a perfect crystal is 0.

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13
Q

Units for entropy (S)

A

J K-1 mol-1

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14
Q

Why does this equation not represent an enthalpy of formation?

A

2 mol of product (must be 1 mol)

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15
Q

Why does this equation not represent an enthalpy of formation?

A

Br is not in its standard (liquid) state

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16
Q

What sign (+ or −) will ∆G° have for a spontaneous reaction?

A

Negative (−)

17
Q

What must be done before we can plug ∆S° into the equation

∆G° = ∆H° − T∆S°?

A

∆S° must be divided by 1000 (to convert to kJ K-1 mol-1)

18
Q

No temperature is given in the question but we must use the equation ∆G° = ∆H° − T∆S°.

What temperature should be used?

A

298 K

(° implies standard conditions)

19
Q

How do we use the equation ∆G° = ∆H° − T∆S° to work out T when the reaction becomes feasible?

A

Let ∆G° = 0

(then T = ∆H° / ∆S°)