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atroni [7]
3 years ago
5

When the oxide of generic metal M is heated at 25C, only a negligible amount of M is produced. MO2(s) <---> M(s)+O2(g) del

ta G = 288.5 kJ/mol
1.) When the reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(S) _________________________________
2.) What is the thermodynamic equilibrium constant for the coupled reaction? K = ????
Chemistry
1 answer:
Valentin [98]3 years ago
3 0

Answer:

1. MO_2(s)+C(s)CO_2(g)+M(s)

2. K=3.57x10^{18}

Explanation:

Hello,

1.) By coupling the given reaction with the formation of carbon dioxide, one states the total reaction as:

MO_2(s)M(s)+O_2(g)\\C(s)+O_2(g)CO_2(g)

____________________________________

MO_2(s)+C(s)CO_2(g)+M(s)

2.) Now, since we know that the Gibbs free energy for the decomposition of the metal is 288.5kJ/mol and the Gibbs free energy for the formation of carbon dioxide has a value of −394.39kJ/mol, the total Gibbs free energy for this process is:

ΔG^o=288.5kJ/mol-394.39kJ/mol=-105.89kJ/mol

So the equilibrium constant is:

K=exp(-\frac{DeltaG^0}{RT} )\\K=exp(-\frac{-105890J/mol}{8.314J/molK*298.15K} )\\K=3.57x10^{18}

Best regards.

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12. What is the volume of 0.07 mol of neon gas at STP?
scoundrel [369]
<h3>Answer:</h3>

2 L Ne

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

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

STP - 22.4 L per mole

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})
  2. Multiply:                              \displaystyle 1.568 \ L \ Ne

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

1.568 L Ne ≈ 2 L Ne

7 0
3 years ago
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