Question

If \(K_p\) for the reaction \(A(g) + 2B(g) \rightleftharpoons 3C(g) + D(g)\) is \(0.05\) atm at \(1000\,K\), its \(K_c\) in terms of \(\frac{x \times 10^{-5}}{R}\). Find \(x\).

Hint:

Always compute \(\Delta n = n_{\text{gaseous products}} - n_{\text{gaseous reactants}}\) carefully before using \(K_p = K_c(RT)^{\Delta n}\).

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
The relationship between the equilibrium constants \(K_{p}\) (expressed in terms of partial pressures) and \(K_{c}\) (expressed in terms of molar concentrations) depends on the change in the number of moles of gaseous products and reactants (\(\Delta n_{g}\)).
Step 2: Key Formula or Approach:
The relation is given by:
\[ K_{p} = K_{c}(RT)^{\Delta n_{g}} \]
Where:
\(\Delta n_{g} = (\text{Total moles of gaseous products}) - (\text{Total moles of gaseous reactants})\)
: Detailed Explanation:
1. Calculate \(\Delta n_{g}\) for the reaction \(A(g) + 2B(g) \rightleftharpoons 3C(g) + D(g)\):
\[ \Delta n_{g} = (3 + 1) - (1 + 2) = 4 - 3 = 1 \]
2. Substitute the given values into the formula:
\[ 0.05 = K_{c}(R \times 1000)^{1} \]
3. Rearrange to find \(K_{c}\):
\[ K_{c} = \frac{0.05}{1000R} \]
\[ K_{c} = \frac{5 \times 10^{-2}}{10^{3}R} \]
\[ K_{c} = \frac{5 \times 10^{-5}}{R} \]
4. Comparing this with the given form \(\frac{x \times 10^{-5}}{R}\), we find:
\[ x = 5 \]
Step 3: Final Answer:
The value of \(x\) is \(5\).