Chemistry 30 Thermodynamics: Entropy Worksheet

Predict whether entropy is increasing (ΔS > 0) or decreasing (ΔS < 0)?
Give a reason for your answer.

steam condenses to water

solid CO₂ sublimes (changes directly from a solid to a gas)

N₂O_{4 (g)} → 2 NO_{2 (g)}

water is heated from 25°C to 50°C.

C₆H_{6 (l)} + O_{2 (g)} → 6 CO_{2 (g)} + 3 H₂O_(l)

Answers:

1. steam condenses to water
	ΔS < 0; entropy decreases because liquid water is less random than gaseous water (steam)
2. solid CO₂sublimes
	ΔS > 0; entropy increases because gaseous CO₂ is more random than the solid state.
3. N₂O_{4 (g)} → 2 NO_{2 (g)}
	ΔS > 0; entropy increases because two moles of a gas are more random than 1 mole of a gas.
4. water is heated from 25°C to 50°C
	ΔS > 0; entropy increases as particles move about more when heated.
5. C₆H_{6 (l)} + O_{2 (g)} → 6 CO_{2 (g)} + 3 H₂O_(l)
	ΔS > 0; entropy increases because there are more moles on the product side of the equation (9 vs 2)

Using a Table of Thermochemical Data, calculate ΔS for the following reaction. Is entropy increasing or decreasing? Is the system becoming more random or less random? Based on entropy changes only, would you predict the reaction to be spontaneous or not?

Na_(s) + ½Cl_{2 (g)} → NaCl_(s)

Look up S° values for all reaction participants, taking care to look in the correct column of the table of thermochemical data. Multiply these values by any coefficients in the balanced equation, then simplify each side of the equation:

Na	+	½ Cl₂	→	NaCl
51.2		½ × 223.1		72.1

162.8				72.1

Use the following formula to find ΔS°

ΔS	= ΣS_products - ΣS_reactants
	= 72.1 - 162.8
	= -90.7 J/K

A negative sign for ΔS° tells us that entropy is decreasing. The system becomes less random (or more ordered). On the basis of entropy changes only we would not expect this reaction to occur spontaneously.

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