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JulsSmile [24]
3 years ago
10

Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo

rable: using the product immediately in the next step, maintaining a high starting-material concentration, or keeping a high product concentration? View Available Hint(s) Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favorable: using the product immediately in the next step, maintaining a high starting-material concentration, or keeping a high product concentration? maintaining a high starting-material concentration keeping a high product concentration using the product immediately in the next step and maintaining a high starting-material concentration using the product immediately in the next step
Chemistry
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature

Q is the reaction quotient

To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

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A) \frac{1mol(SCN)_{2}}{2molNaSCN}

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2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄

A) A conversion factor could be \frac{1mol(SCN)_{2}}{2molNaSCN} , as it has the units that we want to <u>convert to in the numerator</u>, and the units that we want to <u>convert from in the denominator</u>.

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C) With 4 moles of NaSCN and 3 moles of MnSO₄, the<em> reactant </em>is NaSCN so we use that value to calculate the moles of product formed:

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1. Consider the unbalanced equation:
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Answer:

1.) 17.76 moles H₂O

2.) 323.1 g HCN

Explanation:

1.) An equation is balanced when there is an equal amount of each element on both sides of the reaction. The quantities can be modified by adding coefficients in front of the molecules.

The unbalanced equation:

Al₄C₃(s) + H₂O -----> Al(OH)₃(s) + CH₄(g)

<u>Reactants</u>: 4 aluminum, 3 carbon, 2 hydrogen, 1 oxygen

<u>Products</u>: 1 aluminum, 1 carbon, 7 hydrogen, 3 oxygen

The balanced equation:

Al₄C₃(s) + 12 H₂O -----> 4 Al(OH)₃(s) + 3 CH₄(g)

<u>Reactants</u>: 4 aluminum, 3 carbon, 24 hydrogen, 12 oxygen

<u>Products</u>: 4 aluminum, 3 carbon, 24 hydrogen, 12 oxygen

Now that the equation is balanced, you can use the relevant coefficients to construct a mole-to-mole ratio. This will allow you to convert between moles Al₄C₃ to moles H₂O.

 1.48 moles Al₄C₃           12 moles H₂O
----------------------------  x  ------------------------  =  17.76 moles H₂O
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2.) To find the mass of HCN, you need to (1) convert grams NH₃ to moles NH₃ (via molar mass), then (2) convert moles NH₃ to moles HCN (via mole-to-mole from equation coefficients), and then (3) convert moles HCN to grams HCN (via molar mass). It is important to arrange the coefficients in a way that allows for the cancellation of units.

Molar Mass (NH₃): 14.009 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.033 g/mol

Molar Mass (HCN): 1.008 g/mol + 12.011 g/mol + 14.009 g/mol

Molar Mass (HCN): 27.028 g/mol

2 CH₄(g) + 3 O₂(g) + 2 NH₃(g)  ------> 2 HCN(g) + 6 H₂O(g)

203.6 g NH₃            1 mole            2 moles HCN           27.028 g
---------------------  x  ----------------  x  ------------------------  x  -----------------  =
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=  323.1 g HCN

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