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Ne4ueva [31]
3 years ago
9

For a particular reaction, ΔH∘=20.1 kJ/mol and Δ????∘=45.9 J/(mol⋅K). Assuming these values change very little with temperature,

at what temperature does the reaction change from nonspontaneous to spontaneous in the forward direction?T= K Is the reaction in the forward direction spontaneous at temperatures greater than or less than the calculated temperature?
Chemistry
1 answer:
tia_tia [17]3 years ago
7 0

Answer:

The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous

Explanation:

<u>Step 1:</u> Data given

ΔH∘=20.1 kJ/mol

ΔS is 45.9 J/K

<u>Step 2:</u> When is the reaction spontaneous

Consider temperature and pressure = constant.

The conditions for spontaneous reactions are:

ΔH <0

ΔS  > 0

ΔG <0  The reaction is spontaneous at all temperatures

ΔH <0

ΔS  <0

ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)

ΔH >0

ΔS  >0

ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)

<u>Step 3:</u> Calculate the temperature

ΔG <0 = ΔH - T*ΔS

T*ΔS > ΔH

T > ΔH/ΔS

In this situation:

T > (20100 J)/(45.9 J/K)

T > 437.9 K

T > 164.75 °C

The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous

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Explanation:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>2</em><em>×</em><em>1</em><em>cm</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>4</em><em>×</em><em>3</em><em>=</em><em>1</em><em>2</em><em>)</em>

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<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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