Consider the following reaction where K_c = 1.80 times 10^-2 at 698 K: 2HI (g) r

Question

Consider the following reaction where K_c = 1.80 times 10^-2 at 698 K: 2HI (g) rightleftharpoons H_2(g) + I_2(g) A reaction mixture was found to contain 0.284 moles of HI (g), 2.04 times 10^-2 moles of H_2 (g), and 4.31 times 10^-2 moles of I_2 (g), in a 1.00 liter container. Indicate True (T) or False (F) for each of the following: In order to reach equilibrium HI (g) must be produced. In order to reach equilibrium K_c must increase. In order to reach equilibrium H_2 must be produced. Q_c is less than K_c. The reaction is at equilibrium. No further reaction will occur.

Explanation / Answer

Given

Kc = 1.8 * 10-2

HI = 0.284 moles

H2 = 2.04 * 10-2 moles

I2 =  4.31 * 10-2 moles

Volume = 1 L

[HI] = 0.284 mole/L

[H2 ]= 2.04 * 10-2 mole/L

[I2 ]=  4.31 * 10-2 mole/L

Qc = [H2 ]*[I2 ]/[HI]2 = (4.31 * 10-2 mole/L * 2.04 * 10-2 mole/L) / (0.284 mol/L)2 = 1.09 * 10-2

Q < Kc

when Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

so HI will react to produce H2 and I2 to reach equilibrium and kc will remain constant

1. False

2. False

3. True

4. True

5. False

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