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3 years ago

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A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adeno

sine diphosphate (ADP) as described by ATP(aq) + H2O(l)  ADP(aq) + HPO4-2(aq) for which ΔG°rxn = –30.5 kJ/mol at 37.0°C. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 1.2 x10-2 M, [ADP] = 8.4 x10-3 M, and [HPO4-2] = 2.1x10-3 M.

1 answer:

Archy [21]3 years ago

5 0

Answer:

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

K = 1.47 x $10^{-3}$ M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5 $10^{3}$ ) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x $10^{3}$ ) + (-16810.68)

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

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b) The value of k is:

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Acetone (M) Br2 (M) H+ (M) Rate (M/s)

0.30 0.050 0.050 5.7 x 10-5

0.30 0.10 0.050 5.7 x 10-5

0.30 0.050 0.10 1.2 x 10-4

0.40 0.050 0.20 3.1 x 10-4

0.40 0.050 0.050 7.6 x 10-5

A generic rate law for this reaction could be written as follows:

rate = k[Acetone]ᵃ[Br₂]ᵇ[H⁺]ⁿ

The rate for the reaction in trial 2 is:

rate 2 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ

For the reaction in trial 1:

rate 1 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

If we divide both expressions, we can obtain "b": rate2 / rate1:

rate2/rate1 = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ / k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

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If we now take the expressions from trial 3 and 1 and divide them, we can obtain "n":

rate 3/rate 1 = k(0.3)ᵃ(0.050)⁰(0.01)ⁿ/ k(0.3)ᵃ(0.050)⁰(0.050)ⁿ

2.1 = 2ⁿ Applying ln to both side of the equation:

ln 2.1 = n ln2

ln2.1/ln2 = n

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Taking now the reaction in trial 5 and 1 and dividing them:

rate 5/rate 1 = k(0.4)ᵃ(0.050)⁰(0.050) / k(0.3)ᵃ(0.050)⁰(0.050)

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a)Then the rate law can be written as follows:

rate = k[Acetone][Br₂]⁰[H⁺]

It might be suprising that the rate of bromination of acetone does not depend on the concentration of Br₂. However, looking at the reaction mechanism, you can find out why.

b) Now, we can find the constant k for every experiment and calculate its average value:

rate / [Acetone][Br₂]⁰[H⁺] = k

For reaction 1:

k1 = 5.7 ×10⁻⁵M/s / (0.3 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 2: k2 = 5.7 ×10⁻⁵M/s / (0.30 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 3: k3 = 1.2 ×10⁻⁴M/s / (0.30 M)(0.10 M) = 4.0 ×10⁻³ M⁻¹ · s⁻¹

Reaction 4: k4 = 3.1 ×10⁻⁴M/s / (0.40 M)(0.20 M) = 3.9 ×10⁻³ M⁻¹ · s⁻¹

Reaction 5: k5 = 7.6 ×10⁻⁵M/s / (0.4 M)(0.05 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Averge value of k:

k = (k1 + k2 + k3 + k4 + k5)/5 = 3.86 × 10⁻³ M⁻¹ · s⁻¹

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