Proteins bind to metal ions using various amino acids. One such amino acid is cysteine, which has an S
-
at the end of a carbon chain. Typically, four cysteine units use their sulfides to bind tight

Question

Proteins bind to metal ions using various amino acids. One such amino acid is cysteine, which has an S
-
 ﻿at the end of a carbon chain. Typically, four cysteine units use their sulfides to bind tightly to Zn
(
II
)
 ﻿ions. The equilibrium dissociation constant for binding of Zn 
(
II
)
 ﻿at pH 
7
.
0
 ﻿is 
8
.
0
E
-
1
2
.
 ﻿If the delta H for the reaction is 
2
6
.
8
kJ
/
mol
,
 ﻿what is the value of delta S for the reaction at 
2
5
 ﻿degrees C
?        provide step by step answer with proper explaination

VIPSolutions · Accepted Answer

Let's carefully analyze and solve the problem step-by-step.

---

### **Given Data:**

- Equilibrium dissociation constant: $ K_d = 8.0 \times 10^{-12} $
- Temperature: $ T = 25^\circ C $
- $ \Delta H = 26.8\, \text{kJ/mol} $
- We are asked to find: $ \Delta S $ (entropy change) at $ 25^\circ C $

---

### **Step 1: Convert temperature to Kelvin**

$$
T(K) = 25 + 273.15 = 298.15\, K
$$

---

### **Step 2: Recall the relationship between $ K $ and $ \Delta G $**

The standard Gibbs free energy change $ \Delta G^\circ $ relates to the equilibrium constant $ K $ via:

$$
\Delta G^\circ = -RT \ln K
$$

Note: Since the problem states the dissociation constant $ K_d $, which is for the dissociation reaction, the binding equilibrium constant $ K_b $ would be:

$$
K_b = \frac{1}{K_d}
$$

But generally, for binding processes, $ K $ refers to the association constant, so:

$$
K_{assoc} = \frac{1}{K_d}
$$

Given $ K_d = 8.0 \times 10^{-12} $, then:

$$
K_{assoc} = \frac{1}{8.0 \times 10^{-12}} = 1.25 \times 10^{11}
$$

---

### **Step 3: Calculate $ \Delta G^\circ $**

Using the association constant $ K_{assoc} $:

$$
\Delta G^\circ = -RT \ln K_{assoc}
$$

Where:
- $ R = 8.314\, \text{J/(mol·K)} $
- $ T = 298.15\, K $
- $ \ln K_{assoc} = \ln (1.25 \times 10^{11}) $

Calculate $ \ln K_{assoc} $:

$$
\ln(1.25 \times 10^{11}) = \ln(1.25) + \ln(10^{11}) = 0.2231 + 11 \times \ln 10
$$

Since $ \ln 10 \approx 2.3026 $:

$$
\ln K_{assoc} = 0.2231 + 11 \times 2.3026 = 0.2231 + 25.3286 = 25.5517
$$

Now, calculate $ \Delta G^\circ $:

$$
\Delta G^\circ = - (8.314\, \text{J/(mol·K)}) \times 298.15\, K \times 25.5517
$$

$$
\Delta G^\circ = -8.314 \times 298.15 \times 25.5517
$$

Calculate step-by-step:

- $ 8.314 \times 298.15 \approx 2478.8\, \text{J/mol} $
- $ 2478.8 \times 25.5517 \approx 63,366\, \text{J/mol} $

Thus,

$$
\boxed{
\Delta G^\circ \approx -63,366\, \text{J/mol} \approx -63.37\, \text{kJ/mol}
}
$$

---

### **Step 4: Use the thermodynamic relation**

The relation between $ \Delta G^\circ $, $ \Delta H $, and $ \Delta S $:

$$
\Delta G^\circ = \Delta H - T \Delta S
$$

Rearranged to solve for $ \Delta S $:

$$
\Delta S = \frac{\Delta H - \Delta G^\circ}{T}
$$

Convert $ \Delta H $ to Joules:

$$
\Delta H = 26.8\, \text{kJ/mol} = 26,800\, \text{J/mol}
$$

Substitute known values:

$$
\Delta S = \frac{26,800\, \text{J/mol} - (-63,366\, \text{J/mol})}{298.15\, K}
$$

$$
\Delta S = \frac{26,800 + 63,366}{298.15} = \frac{90,166}{298.15}
$$

Calculate:

$$
\Delta S \approx 302.75\, \text{J/(mol·K)}
$$

---

### **### **Final answer:**

$$
\boxed{
\Delta S \approx \mathbf{303\, J/(mol \cdot K)}
}
$$

**This positive entropy change indicates increased disorder during the binding process.**

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