Question

At $527^{\circ} C$, the reaction given below has $K _{c}=4$
$NH _{3}( g ) \rightleftharpoons \frac{1}{2} N _{2}( g )+\frac{3}{2} H _{2}( g )$
What is the $K _{ p }$ for the reaction?
$N _{2}( g )+3 H _{2}( g ) \rightleftharpoons 2 NH _{3}( g )$

• A. $16 \times(800 R )^{2}$
• B. $\left(\frac{800 R }{4}\right)^{-2}$
• C. $\left(\frac{1}{4 \times 800 R }\right)^{2}$
• D. None of these

Answer 1

Answer: (C)

If, $K_{c}=\frac{\left[N_{2}\right]^{1 / 2}\left[H_{2}\right]^{3 / 2}}{\left[N H_{3}\right]^{1}}=4$

then $\frac{\left[ NH _{3}\right]^{2}}{\left[ N _{2}\right]\left[ H _{2}\right]^{3}}=(4)^{-2}=\frac{1}{16}$

also $K_{p}=K_{c} \cdot(R T)^{\Delta n_{g}}$