Question

For the reaction $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, the relation between $K_p$ and $K_c$ is

• A. $K_p = K_c (RT)^{-2}$
• B. $K_p = K_c (RT)^2$
• C. $K_p = K_c (RT)^{-1}$
• D. $K_p = K_c (RT)$

Hint:

Remember, when the change in moles of gas ($\Delta n_g$) is negative, the exponent on $RT$ will also be negative. Conversely, if $\Delta n_g$ is positive, the exponent will be positive.

Answer 1

The Correct Option is A

Step 1: Know the two constants.
For a gas reaction $K_p$ uses partial pressures and $K_c$ uses concentrations. They are linked by the change in number of gas moles.
Step 2: Write the link formula.
The link is\[ K_p = K_c (RT)^{\Delta n_g} \]where $\Delta n_g$ is moles of gas products minus moles of gas reactants.
Step 3: Read the reaction.
The reaction is $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$. On the product side there are 2 moles of gas. On the reactant side there are $1+3 = 4$ moles of gas.
Step 4: Find delta n.
\[ \Delta n_g = 2 - 4 = -2 \]So the gas moles drop by two.
Step 5: Put the value in.
Place $\Delta n_g = -2$ into the link formula to get\[ K_p = K_c (RT)^{-2} \]
Step 6: Final choice.
So the correct relation is $K_p = K_c (RT)^{-2}$.\[ \boxed{K_p = K_c (RT)^{-2}} \]