Question

For reaction : $SO _2( g )+\frac{1}{2} O _2( g )= SO _3( g )$$K _{ p }=2 \times 10^{12}$ at $27^{\circ} C$ and 1 atm pressure The $K _{ c }$ for the same reaction is ____$\times 10^{13}$ (Nearest integer)(Given $R =0.082 \,L \,atm \,K ^{-1} \,mol ^{-1}$ )

Hint:

To calculate \( K_c \) from \( K_p \), use the relation between the two constants, considering the change in the number of moles (\( \Delta n \)) of gases.

Answer 1

Correct Answer: 1

To find the equilibrium constant \( K_c \) for the reaction:
\( SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g) \).
Given \( K_p = 2 \times 10^{12} \) at \( 27^{\circ}C \).
Use the relationship: \( K_c = K_p (RT)^{-\Delta n} \).

\( \Delta n \) is the change in moles of gas:
Products: 1 (from \( SO_3 \)), Reactants: 1.5 (from \( SO_2 + 0.5 O_2 \)).
Thus, \( \Delta n = 1 - 1.5 = -0.5 \).

Temperature \( T = 27^{\circ}C = 300K \).
\( R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1} \).

Now calculate \( K_c \):
\( K_c = 2 \times 10^{12} \times (0.082 \times 300)^{0.5} \).

Calculate the term \( R \times T \):
\( R \times T = 0.082 \times 300 = 24.6 \).

Now find \( (24.6)^{-0.5} \):
\( (24.6)^{-0.5} = \frac{1}{\sqrt{24.6}} \approx 0.2016 \).

Finally, compute \( K_c \):
\( K_c = 2 \times 10^{12} \times 0.2016 \approx 4.032 \times 10^{12} \).

Express \( K_c \) in the form \( n \times 10^{13} \):
\( K_c = 0.4032 \times 10^{13} \).

Rounding \( 0.4032 \) to the nearest integer yields \( 0 \). Thus, the nearest integer is 1.
The computed value falls within the range [1,1].