Question

At 1000 K, the value of $K_c$ for the below reaction is $10 \text{ mol L}^{-1}$. Value of $K_p$ (in atm) is (given $R = 0.082 \text{ atm L mol}^{-1} \text{K}^{-1}$)
${\{A(g) <=> B(g) + C(g)}$

• A. $82$
• B. $0.82$
• C. $8.2$
• D. $820$

Hint:

Conversion between $K_c$ and $K_p$. \[ K_p = K_c(RT)^\Delta n \] Where $\Delta n$ is the change in moles of gas (products - reactants). Use appropriate units.

Answer 1

The Correct Option is D

Use the relation: \[ K_p = K_c(RT)^{\Delta n} \] Here, $\Delta n = 2 - 1 = 1$, $K_c = 10$, $T = 1000~\text{K}$, $R = 0.082~\text{L atm mol}^{-1} \text{K}^{-1}$ \[ K_p = 10 \times (0.082 \times 1000) = 10 \times 82 = 820~\text{atm} \]

Answer 2

Step 1: Understanding the relation between \(K_c\) and \(K_p\)
The equilibrium constants \(K_c\) and \(K_p\) are related by the equation:
\[ K_p = K_c(RT)^{\Delta n} \] where \(R\) is the gas constant, \(T\) is temperature in Kelvin, and \(\Delta n\) is the change in moles of gas.

Step 2: Calculate \(\Delta n\)
For the reaction: \(\mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} + \mathrm{C(g)}\)
Number of moles of gaseous products = 1 (B) + 1 (C) = 2
Number of moles of gaseous reactants = 1 (A)
\[ \Delta n = 2 - 1 = 1 \]

Step 3: Calculate \(K_p\)
Given: \(K_c = 10\, \mathrm{mol\,L^{-1}}\), \(R = 0.082\, \mathrm{atm\,L\,mol^{-1}K^{-1}}\), \(T = 1000\,K\)
\[ K_p = K_c (RT)^{\Delta n} = 10 \times (0.082 \times 1000)^1 = 10 \times 82 = 820\, \mathrm{atm} \]

Step 4: Conclusion
Therefore, the value of \(K_p\) at 1000 K for the given reaction is 820 atm.