Question

$\log \frac{K_p}{K_c}+\log RT=0$ is true relationship for the following reaction:

• A. $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$
• B. $2S{O}_2+O_2\rightleftharpoons 2S{O}_3$
• C. $N_2+3H_2\rightleftharpoons 2N{H}_3$
• D. $\left(2\right)$ and $\left(3\right)$ both

Answer 1

Answer: (B)

For the general reaction, $aA+bB\leftrightharpoons cC+dD$
the relationship between two equilibrium constants is: $K_P=K_C(RT{)}^{\Delta n}$
where, $\Delta n$ (Total moles of products side) - (Total moles of reactants on the reactant side). Hence $\Delta n$= (d+c) - (a+b). R is the gas constant found in the ideal gas law (0.0821 liter x Atm/mole/kelvin) T is the temperature of reaction, kelvin.
This we can use in a relationship for the reaction:
$log\frac{K_P}{K_C}+logRT=0$
$log\frac{K_C(RT{)}^{\Delta n}}{K_C}+logRT=0$
$log\left(\right(RT{)}^{\Delta n}\times \left(RT\right)=0$
$log(RT{)}^{\Delta n+1}=0$
$(RT{)}^{\Delta n+1}=0$
R is a constant and T is the temperature of reaction, so the product cant be zero. Thats why $\Delta n+1=0$ and $\Delta n=-1$
so our relationship is true for the reaction (B) $2S{O}_2+O_2\leftrightharpoons 2S{O}_3$
Because $\Delta n$ for this reaction is
$\Delta n$ =2-2-1= -1
conclusion: hence option (B) is the correct answer.