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Lubov Fominskaja [6]

4 years ago

7

A sample is decomposed in the laboratory and found that it contains 52.2% carbon (C), 13.0% hydrogen (H), and 34.8% oxygen (O).

What is the empirical formula of this compound?

1 answer:

Andrej [43]4 years ago

7 0

The empirical formula would be C2H6O

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Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi

sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Entropy change of the reaction ; ΔS

$\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]$

$=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

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