In preparation of `CaO` from `CaCO_(3)` using the equilibrium, `CaCO_(3)(g)hArrCaO(s)+CO_(2)(g)K_(p)` is expressed as `log K_(p)=7.282-(8500)/(T)`, where T is in Kelvin.
What is value of `K_(p)` at a temperature of 1167 K?

Options

A. 10 atm
B. 2 atm
C. 1 atm
D. zero

Answer

To find the value of \( K_p \) at a temperature of 1167 K, we will follow these steps:

### Step 1: Write down the equation for \( \log K_p \)
The equation given in the problem is:
\[ \log K_p = 7.282 - \frac{8500}{T} \]

### Step 2: Substitute the temperature into the equation
We need to substitute \( T = 1167 \) K into the equation:
\[ \log K_p = 7.282 - \frac{8500}{1167} \]

### Step 3: Calculate the value of \( \frac{8500}{1167} \)
Now, we will perform the division:
\[ \frac{8500}{1167} \approx 7.277 \]

### Step 4: Substitute the calculated value back into the equation
Now we substitute this value back into the equation:
\[ \log K_p = 7.282 - 7.277 \]

### Step 5: Calculate \( \log K_p \)
Now we perform the subtraction:
\[ \log K_p \approx 0.005 \]

### Step 6: Convert from logarithmic form to exponential form
To find \( K_p \), we convert from logarithmic form:
\[ K_p = 10^{\log K_p} = 10^{0.005} \]

### Step 7: Calculate \( K_p \)
Using a calculator:
\[ K_p \approx 1.0116 \]

### Final Answer
Thus, the value of \( K_p \) at 1167 K is approximately:
\[ K_p \approx 1.0116 \text{ (approximately 1)} \]