Question

For the equilibrium, $2A(g)\rightleftharpoons 2B(g)+C(g)$, the value of the equilibrium constant $K_p$ is $0.1662$ atm at $1000$ K. The value of $K_c$ for the equilibrium at the same temperature is $(R=0.0831$ L atm mol$^{-1}$ K$^{-1})$

• A. $1\times10^{-3}$ mol L$^{-1}$
• B. $2\times10^{-3}$ mol L$^{-1}$
• C. $0.4\times10^{-3}$ mol L$^{-1}$
• D. $1.6\times10^{-3}$ mol L$^{-1}$
• E. $0.6\times10^{-3}$ mol L$^{-1}$

Hint:

If $\Delta n$ is positive, then $K_p>K_c$ usually because gas moles increase.

Answer 1

The Correct Option is B

Concept: Chemistry - Relation between $K_p$ and $K_c$. [ K_p=K_c(RT)^{\Delta n} ]
Step 1: Find change in moles. For reaction: [ 2A(g)\rightleftharpoons 2B(g)+C(g) ] Reactant moles = 2 Product moles = 3 [ \Delta n=3-2=1 ]
Step 2: Use formula. [ K_p=K_c(RT)^1 ] [ 0.1662=K_c(0.0831\times1000) ]
Step 3: Calculate $K_c$. [ 0.0831\times1000=83.1 ] [ K_c=\frac{0.1662}{83.1}=0.002 ] [ K_c=2\times10^{-3} ]
Step 4: Final answer. [ \boxed{2\times10^{-3}\text{ mol L}^{-1}} ]
Hence, correct option is (B).