Properties of the Equilibrium Constant Practice

Sample questions

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1. Sample 1difficulty 2/5

   If K_fwd = 4, K_rev =

   • A

     0.25

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   • B

     4

   • C

     16

   • D

     −4

   Why

   K for reverse = 1/K_fwd = 1/4 = 0.25.

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2. Sample 2difficulty 3/5

   K_p and K_c are related by

   • A

     K_p = K_c/T

   • B

     K_p = K_c(RT)^Δn

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   • C

     K_p = K_c·R·T

   • D

     K_p = K_c·n

   Why

   Δn = (moles gas products) − (moles gas reactants).

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3. Sample 3difficulty 4/5

   A ⇌ B has K=2; B ⇌ C has K=3. K for A ⇌ C is

   • A

     1

   • B

     6

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   • C

     5

   • D

     0.67

   Why

   Add reactions → multiply K's: 2 × 3 = 6.

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4. Sample 4difficulty 4/5

   At 500 K, $K_p$ = 1.5 for $N_2O_4(g) \rightleftharpoons 2NO_2(g)$. What is $K_c$? (R = 0.08206 L·atm/mol·K)

   • A

     $K_c = 0.244$

   • B

     $K_c = 0.0366$

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   • C

     $K_c = 1.50$

   • D

     $K_c = 61.5$

   Why

   $K_p = K_c(RT)^{\Delta n}$ with $\Delta n$ = 2-1 = 1. $K_c$ = $K_p/(RT)$ = 1.5/((0.08206)(500)) = 1.5/41.03 = 0.0366.

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5. Sample 5difficulty 4/5

   For 2 NO$_2$(g) ⇌ N$_2$O$_4$(g), Δn(gas) = 1 − 2 = −1. The relation between Kp and Kc is

   • A

     Kp = Kc

   • B

     Kp = Kc$^2$

   • C

     Kp = Kc(RT)$^{Δn}$ = Kc/(RT)

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   • D

     Kp = Kc × RT

   Why

   $K_p = K_c (RT)^{\Delta n}$ where Δn = mol gas products − mol gas reactants. Here Δn = −1, so $K_p = K_c/(RT)$.

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