To solve this problem, we need to follow these steps:

1. **Write the balanced chemical equation for the Claus process:**
$$ 2H_2S(g) + O_2(g) \rightarrow 2S(s) + 2H_2O(g) $$

2. **Use the ideal gas law to find the moles of dioxygen consumed per second:**
The ideal gas law is given by:
$$ PV = nRT $$
Where:
- $ P $ is the pressure (0.82 atm)
- $ V $ is the volume (620 L)
- $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
- $ T $ is the temperature in Kelvin (179°C + 273 = 452 K)

Rearrange the ideal gas law to solve for $ n $ (moles of $ O_2 $):
$$ n = \frac{PV}{RT} $$
$$ n = \frac{(0.82 \text{ atm})(620 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(452 \text{ K})} $$
$$ n \approx 13.6 \text{ mol/s} $$

3. **Determine the moles of sulfur dioxide produced per second:**
From the balanced equation, 1 mole of $ O_2 $ produces 1 mole of $ SO_2 $. Therefore, the moles of $ SO_2 $ produced per second is also 13.6 mol/s.

4. **Convert moles of sulfur dioxide to mass:**
The molar mass of $ SO_2 $ is:
$$ 32.07 \text{ g/mol (S)} + 2 \times 16.00 \text{ g/mol (O)} = 64.07 \text{ g/mol} $$

The mass of $ SO_2 $ produced per second is:
$$ 13.6 \text{ mol/s} \times 64.07 \text{ g/mol} = 871.35 \text{ g/s} $$

5. **Convert grams per second to kilograms per second:**
$$ 871.35 \text{ g/s} = 0.87 \text{ kg/s} $$

Therefore, the rate at which sulfur dioxide is being produced is approximately $ 0.87 \text{ kg/s} $.

Question

To solve this problem, we need to follow these steps:

1. **Write the balanced chemical equation for the Claus process:**
   $$ 2H_2S(g) + O_2(g) \rightarrow 2S(s) + 2H_2O(g) $$

2. **Use the ideal gas law to find the moles of dioxygen consumed per second:**
   The ideal gas law is given by:
   $$ PV = nRT $$
   Where:
   - $ P $ is the pressure (0.82 atm)
   - $ V $ is the volume (620 L)
   - $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
   - $ T $ is the temperature in Kelvin (179°C + 273 = 452 K)

Rearrange the ideal gas law to solve for $ n $ (moles of $ O_2 $):
   $$ n = \frac{PV}{RT} $$
   $$ n = \frac{(0.82 \text{ atm})(620 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(452 \text{ K})} $$
   $$ n \approx 13.6 \text{ mol/s} $$

3. **Determine the moles of sulfur dioxide produced per second:**
   From the balanced equation, 1 mole of $ O_2 $ produces 1 mole of $ SO_2 $. Therefore, the moles of $ SO_2 $ produced per second is also 13.6 mol/s.

4. **Convert moles of sulfur dioxide to mass:**
   The molar mass of $ SO_2 $ is:
   $$ 32.07 \text{ g/mol (S)} + 2 \times 16.00 \text{ g/mol (O)} = 64.07 \text{ g/mol} $$

The mass of $ SO_2 $ produced per second is:
   $$ 13.6 \text{ mol/s} \times 64.07 \text{ g/mol} = 871.35 \text{ g/s} $$

5. **Convert grams per second to kilograms per second:**
   $$ 871.35 \text{ g/s} = 0.87 \text{ kg/s} $$

Therefore, the rate at which sulfur dioxide is being produced is approximately $ 0.87 \text{ kg/s} $.

Knowee AI · Accepted Answer