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Arte-miy333 [17]
3 years ago
9

The following reaction has an equilibrium constant (Keq) of 1 under standard conditions and can be catalysed by the enzyme aspar

tate transaminase: Reaction 1: aspartate + a-ketoglutarate oxaloacetate + glutamate () (ii) (iii) (iv) What is the AGⓇ of this reaction? Which will be favoured - the forward reaction, the reverse reaction, or neither? What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed?
Chemistry
1 answer:
Trava [24]3 years ago
4 0

Answer:

  1. What is the AGⓇ of this reaction? 0.
  2. Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
  3. What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
  4. If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.

Explanation:

1. To calculate the delta G of a reaction given the K, we use the following equation:

ΔG°= -RT ln K.

Which gives us 0 when K is 1.

2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.

3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.

4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.

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The balanced molecular equation for the given reaction is:

2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).

<h3>FURTHER EXPLANATION</h3>

To check that the equation is balanced count how many atoms are present for each element in the reactant and product side. If they are the same before and after reaction for all elements, then the reaction is deemed balanced.

The atom counting for the equation is shown below:

2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).

<u>Reactants </u>                  →                 <u>Products</u>

C:  (2 x 2) =4                                  C: (2 x 2) = 4

H: (2 x 4) + (1 x 2) =10                    H: (3 x 2) + (2 X 2) = 10

O: (2 x 2) + (1 x 2) = 6                    O: (2 x 2) + (2 x 1) = 6

Ba: 1                                               Ba: 1

Since the number of atoms of each element are similar in the reactants and products, the equation is balanced.

To determine the state of the substances, consider their solubility.

The reactants are both aqueous (aq) as indicated in the problem.

The first product, Ba(CH_3COO)_2 is aqueous (aq) because based on the solubility rule, it will dissolve in water. Acetates are generally soluble.

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<h3>LEARN MORE</h3>
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Keywords: acid-base, neutralization, balancing equations, molecular equation

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A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
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Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

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If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

7 0
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The number of moles of solute in 200.ml of a 0.500 m solution is
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Answer:

I believe the answer is 0.100.

Explanation:

Hope my answer has helped you!

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