The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent

Q: Why do we set $\Delta G = 0$?

Setting $\Delta G = 0$ represents the threshold (equilibrium) between spontaneous and non-spontaneous oxidation. To completely prevent oxidation, we need $\Delta G \ge 0$, so the minimum required hydrogen pressure corresponds to the exact point where $\Delta G = 0$.

Q: Why is the partial pressure of water $0.01$ bar?

The problem states the total pressure is $1\text{ bar}$ and water vapor is $1\text{ mole }\%$ of the mixture. According to Dalton's Law, partial pressure equals mole fraction times total pressure. So, $0.01 \times 1\text{ bar} = 0.01\text{ bar}$.

Q: Why are solids excluded from the reaction quotient $Q$?

In thermodynamics, the activity of pure solids and pure liquids is defined as exactly $1$. Therefore, they do not affect the value of the reaction quotient $Q$ or the equilibrium constant $K$.

Q: Why did we multiply $\Delta G^\circ$ by $1000$?

The given $\Delta G^\circ$ values are in kilojoules per mole ($\text{kJ mol}^{-1}$), but the universal gas constant $R$ is given in Joules per Kelvin per mole ($\text{J K}^{-1}\text{mol}^{-1}$). We must multiply by $1000$ to ensure all energy units in the equation match.

Q: How do we calculate $\ln(0.01)$ without a calculator?

You can use the properties of logarithms. Rewrite $0.01$ as $10^{-2}$. Then, $\ln(10^{-2}) = -2 \ln(10)$. Since the problem gives $\ln(10) = 2.3$, you simply multiply $-2$ by $2.3$ to get $-4.6$.

Question

The surface of copper gets tarnished by the formation of copper oxide. $	ext{N}_2$ gas was passed to prevent the oxide formation during heating of copper at $1250	ext{ K}$. However, the $	ext{N}_2$ gas contains $1	ext{ mole }\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below :
$$2	ext{Cu(s)} + 	ext{H}_2	ext{O(g)} ightarrow 	ext{Cu}_2	ext{O(s)} + 	ext{H}_2	ext{(g)}$$
$p_{	ext{H}_2}$ is the minimum partial pressure of $	ext{H}_2$ (in bar) needed to prevent the oxidation at $1250	ext{ K}$. The value of $\ln(p_{	ext{H}_2})$ is ______.
(Given : total pressure $= 1	ext{ bar}$, $R$ (universal gas constant) $= 8	ext{ J K}^{-1}	ext{mol}^{-1}$, $\ln(10) = 2.3$. $	ext{Cu(s)}$ and $	ext{Cu}_2	ext{O(s)}$ are mutually immiscible.
At $1250	ext{ K}$ : $2	ext{Cu(s)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{Cu}_2	ext{O(s)}$; $\Delta G^	heta = -78,000	ext{ J mol}^{-1}$
$	ext{H}_2	ext{(g)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{H}_2	ext{O(g)}$; $\Delta G^	heta = -1,78,000	ext{ J mol}^{-1}$; $G$ is the Gibbs energy)

Sigma Percentile · Accepted Answer

### The Battle Against Oxidation

Imagine you are heating a piece of copper to a scorching $1250	ext{ K}$. At such high temperatures, copper is highly susceptible to oxidation. To protect it, we flush the system with nitrogen gas ($	ext{N}_2$). However, there's a catch—the nitrogen gas isn't perfectly pure; it contains $1	ext{ mole }\%$ of water vapor ($	ext{H}_2	ext{O}$).

This tiny amount of water vapor is enough to act as an oxidizing agent, threatening to tarnish our pristine copper surface according to the following reaction:

$$2	ext{Cu(s)} + 	ext{H}_2	ext{O(g)} ightarrow 	ext{Cu}_2	ext{O(s)} + 	ext{H}_2	ext{(g)}$$

Our mission is to find the minimum partial pressure of hydrogen gas ($p_{	ext{H}_2}$) that we must maintain in the system to completely halt this oxidation process.

### Hess's Law to the Rescue

To understand the thermodynamics of our target reaction, we first need its standard Gibbs free energy change ($\Delta G^\circ$). The problem provides us with the thermodynamic data for two related reactions at $1250	ext{ K}$:

1. The formation of copper oxide:
$$2	ext{Cu(s)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{Cu}_2	ext{O(s)} \quad \Delta G_1^\circ = -78,000	ext{ J mol}^{-1}$$

2. The formation of water vapor:
$$	ext{H}_2	ext{(g)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{H}_2	ext{O(g)} \quad \Delta G_2^\circ = -1,78,000	ext{ J mol}^{-1}$$

We can use Hess's Law to construct our target reaction. By reversing the second reaction (which changes the sign of its $\Delta G^\circ$) and adding it to the first reaction, the oxygen molecules cancel out perfectly!

$$\Delta G^\circ_{	ext{net}} = \Delta G_1^\circ - \Delta G_2^\circ$$
$$\Delta G^\circ_{	ext{net}} = -78,000 - (-1,78,000) = +100,000	ext{ J mol}^{-1}$$

A positive standard free energy tells us that under standard conditions, the oxidation is not spontaneous. But we are not under standard conditions!

### The Equilibrium Condition

The actual spontaneity of the reaction depends on the reaction quotient, $Q$, through the isotherm equation:

$$\Delta G = \Delta G^\circ + RT \ln Q$$

For the oxidation of copper to be prevented, the forward reaction must not be spontaneous. This requires $\Delta G \ge 0$. To find the **minimum** required partial pressure of hydrogen, we look at the boundary condition where the system is exactly at equilibrium, meaning $\Delta G = 0$.

Let's define our reaction quotient, $Q$. Since solids ($	ext{Cu}$ and $	ext{Cu}_2	ext{O}$) have an activity of 1, they do not appear in the expression. $Q$ is simply the ratio of the partial pressures of the gaseous product and reactant:

$$Q = \frac{p_{	ext{H}_2}}{p_{	ext{H}_2	ext{O}}}$$

We are given that the total pressure is $1	ext{ bar}$, and water vapor constitutes $1	ext{ mole }\%$ of the gas mixture. Therefore, the partial pressure of water vapor is:

$$p_{	ext{H}_2	ext{O}} = 0.01 	imes 1	ext{ bar} = 0.01	ext{ bar}$$

### The Final Calculation

Now, we substitute all our known values into the isotherm equation. Remember to use the value of $R = 8	ext{ J K}^{-1}	ext{mol}^{-1}$ as given in the problem, and ensure all energy units are in Joules.

$$0 = 100,000 + (8)(1250) \ln\left(\frac{p_{	ext{H}_2}}{0.01}ight)$$

Let's simplify the arithmetic. Multiplying $8$ by $1250$ gives exactly $10,000$.

$$0 = 100,000 + 10,000 \ln\left(\frac{p_{	ext{H}_2}}{0.01}ight)$$

Dividing the entire equation by $10,000$ makes it much cleaner:

$$-10 = \ln\left(\frac{p_{	ext{H}_2}}{0.01}ight)$$

Using the properties of logarithms, we can expand the right side:

$$-10 = \ln(p_{	ext{H}_2}) - \ln(0.01)$$

We need to evaluate $\ln(0.01)$. We can rewrite $0.01$ as $10^{-2}$:

$$\ln(0.01) = \ln(10^{-2}) = -2 \ln(10)$$

The problem kindly provides $\ln(10) = 2.3$. Substituting this in:

$$\ln(0.01) = -2(2.3) = -4.6$$

Finally, we plug this back into our main equation to solve for $\ln(p_{	ext{H}_2})$:

$$-10 = \ln(p_{	ext{H}_2}) - (-4.6)$$
$$-10 = \ln(p_{	ext{H}_2}) + 4.6$$
$$\ln(p_{	ext{H}_2}) = -10 - 4.6 = \mathbf{-14.6}$$

And there we have it! The natural logarithm of the minimum hydrogen partial pressure required to prevent the tarnishing of copper is **-14.6**.

Enter Numerical Value:

Visualized Solution

$2 Cu(s) + H_{2} O(g) \to Cu_{2} O(s) + H_{2} (g)$

$Given Standard Free Energies$

$Applying Hess’s Law$

$Condition for Non-Oxidation$

$Reaction Quotient (Q)$

$Substituting the Values$

$Simplifying the Equation$

$Final Calculation$

The Sigma Insight: Entropy and Free Energy

The Battle Against Oxidation

Hess's Law to the Rescue

The Equilibrium Condition

The Final Calculation

Similar Questions

The following reaction is performed at $298 K$ $2 NO (g) + O_{2} (g) ⇌ 2 NO_{2} (g)$ The standard free energy of formation of $NO (g)$ is $86.6 kJ/mol$ at $298 K$ . What is the standard free energy of formation of $NO_{2} (g)$ at $298 K$ ? ( $K_{p} = 1.6 \times 1 0^{12}$ )

Comprehension Passage

The value of standard enthalpy, $Δ H^{⊖}$ (in $kJ mol^{- 1}$ ) for the reaction is________.

The value of $Δ S^{⊖}$ (in $J K^{- 1} mol^{- 1}$ ) for the given reaction, at $1000 K$ is________.

For the chemical reaction, $X ⇌ Y$ , the standard reaction Gibbs energy depends on temperature $T$ (in K) as $Δ_{r} G^{\circ} (in kJ mol^{- 1}) = 120 - \frac{3}{8} T$ The major component of the reaction mixture at $T$ is

For the process $H_{2} O (l) \to H_{2} O (g)$ at $T = 10 0^{\circ} C$ and $1$ atmosphere pressure, the correct choice is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Standard Free Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Hess’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Non-Oxidation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction Quotient (Q)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation