Hot copper turnnings can be used as "oxygen getter" for inert gas supplies by slowly passing the gas over the copper tumning at 600 K.
`2Cu_((s)) + (1)/(2) O_(2(g)) hArr Cu_(2) O_((s)) : K_(p) = 7.5 xx 10^(10)`
The number of molecules of `O_(2)` are left in 1 Lof a gas supply after equilibrium has been reached is X.10X is ?

Answer

To solve the problem, we need to determine the number of molecules of \( O_2 \) left in a 1 L gas supply after equilibrium has been reached. We will use the equilibrium constant \( K_p \) and the ideal gas law to find the solution step by step.

### Step 1: Write the balanced chemical equation and the expression for \( K_p \)

The balanced chemical equation is:
\[ 2 \text{Cu}_{(s)} + \frac{1}{2} \text{O}_{2(g)} \rightleftharpoons \text{Cu}_2\text{O}_{(s)} \]

The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{\text{Cu}_2\text{O}}}{(P_{\text{O}_2})^{1/2}} \]

Since \( \text{Cu}_2\text{O} \) is a solid, its pressure does not appear in the expression. Therefore, we can simplify it to:
\[ K_p = \frac{1}{(P_{\text{O}_2})^{1/2}} \]

Given \( K_p = 7.5 \times 10^{10} \), we can rearrange this to find \( P_{\text{O}_2} \):
\[ P_{\text{O}_2}^{1/2} = \frac{1}{K_p} \]
\[ P_{\text{O}_2} = \left(\frac{1}{K_p}\right)^2 \]

### Step 2: Calculate \( P_{\text{O}_2} \)

Substituting the value of \( K_p \):
\[ P_{\text{O}_2} = \left(\frac{1}{7.5 \times 10^{10}}\right)^2 = \frac{1}{(7.5 \times 10^{10})^2} \]
Calculating this gives:
\[ P_{\text{O}_2} = \frac{1}{5.625 \times 10^{21}} \approx 1.77 \times 10^{-22} \text{ atm} \]

### Step 3: Use the ideal gas law to find the number of moles of \( O_2 \)

Using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P = 1.77 \times 10^{-22} \) atm
- \( V = 1 \) L
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 600 \) K

Rearranging for \( n \):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1.77 \times 10^{-22}) \times 1}{0.0821 \times 600} \]
Calculating this gives:
\[ n \approx \frac{1.77 \times 10^{-22}}{49.26} \approx 3.6 \times 10^{-24} \text{ moles} \]

### Step 4: Calculate the number of molecules of \( O_2 \)

Using Avogadro's number \( N_A = 6.022 \times 10^{23} \) molecules/mol:
\[ \text{Number of molecules} = n \times N_A = (3.6 \times 10^{-24}) \times (6.022 \times 10^{23}) \]
Calculating this gives:
\[ \text{Number of molecules} \approx 2.17 \approx 2.2 \text{ molecules} \]

### Step 5: Determine the value of \( 10x \)

From the problem, we have \( X = 2.2 \). Therefore:
\[ 10x = 10 \times 2.2 = 22 \]

### Final Answer
The value of \( 10x \) is \( \boxed{22} \).

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