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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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<u>The answer is </u><u>9.94 ml.</u>

<h3>What is density?</h3>
  • Density is a word we use to describe how much space an object or substance takes up (its volume) in relation to the amount of matter in that object or substance (its mass).
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Given,

        The density of ethanol, C2H5OH = 0.789 g/mL

n (CO_{2} ) = \frac{m}{M}  = \frac{15G}{44 g/mol} } = 0.341 mol;

n ( C_{2} H_{2} OH) = \frac{n (CO_{2}) }{2}  = \frac{0.341}{2}  = 0.1705  mol;

m (C_{2} H_{2} OH) = 0.1705 mol * 46  g/ mol = 7.843 g

V (C_{2} H_{2} OH ) = \frac{7.843}{0.789} = 9.94 ml.

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<u>The complete question is -</u>

If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according to the following chemical equation?

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

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