Chemical Equilibrium Question 6
Question 6 - 31 January - Shift 1
For reaction: $SO_2(g)+\frac{1}{2} O_2(g) \rightarrow SO_3(g)$
$K_P=2 \times 10^{12}$ at $27^{\circ} C$ and $1 atm$ pressure. The $K_c$ for the same reaction is $\quad \times 10^{13}$. (Nearest integer)
(Given $R=0.082 L atm K^{-1} mol^{-1}$ )
Show Answer
Answer: (1)
Solution:
Formula:
The relationship between $K_p$ and $K_c$ for a gaseous reaction is given by the equation:
$$ K_p=K_c(R T)^{\Delta n}…………………..(1) $$
Where:
- $R=0.082 \mathrm{~L} \mathrm{~atm}^{-1} \mathrm{~mol}^{-1}$ is the gas constant,
- $T$ is the temperature in Kelvin,
- $\Delta n$ is the change in moles of gas (products - reactants).
For the reaction:
$$ \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) $$
The change in moles of gas is:
$$ \Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2} $$
Given:
- $K_p=2 \times 10^{12}$
- Temperature, $T=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
Substituting these values into the equation (1):
$$ K_c=\frac{K_P}{(R T)^{\Delta n}}=\frac{2 \times 10^{12}}{(0.082 \times 300)^{-\frac{1}{2}}} = 9.92 \times 10^{12} \approx 1 \times 10^{13} $$
The equilibrium constant $K_c$ for the reaction at $27^{\circ} \mathrm{C}$ is approximately $1 \times 10^{13}$. So, answer is 1.
BETA
