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2 years ago

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for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g) h2o(l) → 2 hno3(l) no(g) j/k

(b) n2(g) 3 f2(g) → 2 nf3(g) j/k (c) c6h12o6(s) 6 o2(g) → 6 co2(g) 6 h2o(g) j/k

1 answer:

mylen [45]2 years ago

8 0

The entropy change, so the appropriate sign is negative. -286.23 J/K (decrease in entropy) Non-spontaneous process.

Entropy is a systematic idea in addition to measurable physical belongings that is most normally associated with a country of ailment, randomness, or uncertainty.

A result of entropy is that certain processes are irreversible or impossible, apart from the requirement of now not violating the conservation of electricity, the latter being expressed inside the first regulation of thermodynamics. Entropy is valuable to the second regulation of thermodynamics.

ENTROPY CHANGE = ENTROPY OF PRODUCT - ENTROPY OF REACTANT

ΔS = ΣΔSf(product) - ΣΔSf(reactant)

1. At STP

[2ΔSf(HNO3 (aq)) + 1ΔSf(NO (g))] - [3ΔSf(NO2 (g)) + 1ΔSf(H2O (ℓ))]

ΔSf(HNO3 (aq) = 146.44 J/K

ΔSf(NO (g) = 210.65 J/K

ΔSf(NO2 (g) = 239.95 J/K

ΔSf(H2O (ℓ) = 69.91 J/K

2(146.44) + 1(210.65)] - [3(239.95) + 1(69.91)] = -286.23 J/K

-286.23 J/K (decrease in entropy)

Non-spontaneous process.

Learn more about entropy here:-brainly.com/question/419265

#SPJ4

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Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

<u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

<u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

<u>For b:</u>

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

<u>For process 1:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

<u>For process 2:</u>

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

<u>For process 3:</u>

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

<u>For process 4:</u>

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

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