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bija089 [108]
3 years ago
7

The standard free-energy changes for the reactions below are given.Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/molATP → ADP

+ Pi ∆ G'° = –30.5 kJ/mol
What is the overall ∆ G'° for the following reaction?
Phosphocreatine + ADP → creatine + ATP
Chemistry
1 answer:
Anton [14]3 years ago
7 0

Answer:

Gibbs free-energy of the reaction = (–12.5 kJ/mol)

Explanation:

The Gibbs free-energy of a reaction predicts the spontaneity or feasibility of a given chemical reaction.

<u>Given the standard Gibbs free energy changes</u>:

Phosphocreatine → creatine + Pi,  ∆G° = –43.0 kJ/mol     ...(1)

ATP → ADP + Pi , ∆G° = –30.5 kJ/mol      ....(2)

<u>Now to calculate the Gibbs free-energy of the given chemical reaction</u>: Phosphocreatine + ADP → creatine + ATP; the <em>equation (2) is reversed</em> to give:

ADP + Pi  → ATP, ∆G° = + 30.5 kJ/mol      ...(3)

<u>Now the equation (3) and (1) are added</u>, to give:

Phosphocreatine + ADP + Pi→ creatine + ATP + Pi

⇒ Phosphocreatine + ADP → creatine + ATP  

 

Therefore, to <u>calculate the Gibbs free-energy of the reaction, the standard Gibbs free energy changes of the equations (1) and (3) are added similarly</u>:

Gibbs free-energy of the reaction: ∆G° = (–43.0 kJ/mol) + ( + 30.5 kJ/mol) = (–12.5 kJ/mol)

<u><em>Therefore, the Gibbs free-energy of the reaction </em></u><u><em>= </em></u><u><em>(–12.5 kJ/mol)</em></u>

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icang [17]
<span>1. rate = k[A][B]² - is the best choice,
because when you double A. </span>rate = k[2A][B]²=rate = k*2*[A][B]²

when you double B . rate = k[A][2B]²=rate = k[A][B]²*2²= rate = k[A][B]²*4
<span>2.
1) rate1 =[0.20]^x*[0.15]^y=2.4*10⁻²
    rate2=[0.20]^x*[0.30]^y=4.8*10⁻²
divide equation of rate2 by rate 1
rate2/rate1     </span>[0.20]^x*[0.30]^y/([0.20]^x*[0.15]^y)=4.8*10⁻²/2.4*10⁻²
<span> 
</span> [0.30]^y/*[0.15]^y=4.8/2.4 , ( [0.30]/[0.15])^y=2, 2^y=1  y =1 , so exponent for [Y] will be 1

2)rate1= [<span>0.20]^x*[0.30]^y = 4.8 × 10⁻²
    rate2=[0.40]^x *[0.30 ]^y= 19.2 × 10⁻²
</span>divide equation of rate2 by rate 1
rate2/rate1     [0.40]^x*[0.30]^y/([0.20]^x*[0.30]^y)=19.2*10⁻²/4.8*10⁻²
                      [0.40]^x/[0.20]^x=19.2/4.8
                      ([0.40]/[0.20])^x= 4, 
                      (2)^x=4, x=2, so  so exponent for [X] will be 2

3) rate=k[X]²[Y]
4) to find k 
take <span> [X]=0.20 M, [Y]= 0.30 M  rate=4.8 × 10⁻²</span> M/min   
 rate=k[X]²[Y]
 4.8 × 10⁻² M/min=k[0.20M]²[0.30M]
4.8 × 10⁻² M/min=k*(0.04*0.3)M³
k=(4.8 × 10⁻² M/min)/(0.012 M³)= 4 min/M²
5) final equation  
rate=(4 min/M²)*k[X]²[Y]

    <span>
    
</span>
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