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Annette [7]
3 years ago
12

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

es of I2(g) react at standard conditions. S°surroundings = J/K
Chemistry
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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Mars2501 [29]

Answer:N

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Explanation:

N

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Let us balance this equation by counting the number of atoms on both sides of the arrow.

N

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N=2 , H=2 N=1, H=3

To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N

H

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N

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H

2

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N=2 , H=2 N=2 , H= 6

To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of

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H

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-----> 2N

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N=2 , H=6 N=2 , H= 6

Answer link

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4 0
3 years ago
Read 2 more answers
20 poll
aleksley [76]

Answer: n=15.56moles

Explanation:

PV = nRT

where  

P is pressure in atmospheres

V is volume in Liters

n is the number of moles of the gas

R is the ideal gas constant = given as (0.0821L -atm/k-mol

PV = nRT

n= PV/RT

n= (1.5 X 230)/ (0.0821  X 270)

n= 15.56 moles

5 0
3 years ago
2C2H2 + 5O2 → 4CO2 + 2H2O
Mumz [18]

Answer:

d. Two moles of carbon dioxide were produced from this reaction

Explanation:

The given chemical reaction can be written as follows;

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

From the above chemical reaction, we have;

Two moles of C₂H₂ reacts with five moles of O₂ to produce four moles of CO₂ and two moles of H₂O

We have;

One mole of C₂H₂ will react with two and half moles of O₂ to produce <em>two moles of CO₂</em> and one mole of H₂O

Therefore, in the above reaction, when one mole of C₂H₂ is used, two moles of CO₂ will be produced.

8 0
3 years ago
Someone please help me with this!!
pantera1 [17]
I would think that it’s C, but try to get a second opinion as well!!
4 0
2 years ago
Identify whether longhand notation or noble-gas notation was used in each case below.
n200080 [17]

Answer:

The given electronic configuration is long hand notation.

Explanation:

Long-hand notation of representing electronic configuration is defined as the arrangement of total number of electrons that are present in an element.

Noble-gas notation of representing electronic configuration is defined as the arrangement of valence electrons in the element. The core electrons are represented as the previous noble gas of the element that is considered.

The given electronic configuration of potassium (K):  

The above configuration has all the electrons that are contained in the nucleus of an element. Thus, this configuration is a long-hand notation.

6 0
3 years ago
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