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Flura [38]
3 years ago
13

What does a process require to be spontaneous at all temperatures? Answer a catalyst and lower activation energy positive enthal

py and negative entropy changes a heat source and high activation energy negative enthalpy and positive entropy changes
Chemistry
2 answers:
tino4ka555 [31]3 years ago
8 0

Answer: negative enthalpy and positive entropy changes

Explanation:

\Delta G=\Delta H-T\Delta S

\Delta G = free energy change

\Delta H = enthalpy change

T= temperature in Kelvin

\Delta S = entropy change

\Delta G= -ve, reaction is spontaneous

If \Delta H=-ve \Delta S=+ve

\Delta G=(-ve)-T(+ve)

\Delta G=(-ve)(-ve)

\Delta G=(-ve)

Thus reaction is spontaneous.

DedPeter [7]3 years ago
7 0
I think the correct answer from the choices listed above is the last option A process requires negative enthalpy and positive entropy changes to be spontaneous at all temperatures. These processes <span>proceed when left to themselves, and in the absence of any attempt to drive them in reverse. Hope this answers the question.</span>

<span />
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<h3>Answer:</h3>

1.69 g Mg₃N₂

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

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

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 1.68873 \ g \ Mg_3N_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

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