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AveGali [126]
2 years ago
13

A certain reaction is spontaneous at temperatures below 400. K but is not spontaneous at temperatures above 400. K. If ΔH° for t

he reaction is -20. kJ mol-1 and it is assumed that ΔH° and ΔS° do not change appreciably with temperature, then the value of ΔS° for the reaction is
Chemistry
1 answer:
jeka57 [31]2 years ago
7 0

Answer:

ΔS = 0.05 kJ /mol or 50 J /mol

Explanation:

As given that the reaction is spontaneous below 400 K and non spontaneous above 400K. Thus the reaction is at equilibrium at 400 K temperature.

At equilibrium the change in free energy ΔG is zero.

The relation between free energy change, enthalpy change, entropy change and temperature is

ΔG = ΔH - TΔS

Where

ΔH = enthalpy change

T = temperature

ΔS = entropy change

at equilibrium (400 K)

ΔG = 0 = (-20) + 400XΔS

ΔS = 0.05 kJ /mol or 50 J /mol

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A 25.0 g sample of metal at
Nadya [2.5K]

The change in temperature of the metal is 6.1°C. Details about change in temperature can be found below.

<h3>How to calculate change in temperature?</h3>

The change in temperature of a substance can be calculated by subtracting the initial temperature of the substance from the final temperature.

According to this question, a 25.0 g sample of metal at 16.0 °C is warmed to 22.1 °C by 259J of energy.

This means that the change in temperature of the metal can be calculated as:

∆T = 22.1°C - 16°C

∆T = 6.1°C

Therefore, the change in temperature of the metal is 6.1°C.

Learn more about change in temperature at: brainly.com/question/19051558

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