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3 years ago

Given the following unbalanced equation:

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

7 0

<h3>Answer:</h3>

11.84 mol CoF₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 11.84 \ mol \ CoF_2$

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svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG = 1.8 x 10⁴ J

K = 8.2 x 10⁻²

Explanation:

a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂

ΔHf kJ/mol -12.5 103.8 0

ΔGºf kJ/K 119.7 202.5 0

Sº J/mol 255 238 130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate ΔHºrxn we have

ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn = ΔHrxn -TΔS

we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

ΔGrxn = ΔHrxn -TΔS

to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.

ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K

ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²

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I love kay.

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