Question

Consider the following reaction where Kc = 1.80×10-2 at 698 K:2 HI (g) H2 (g) + I2 (g)A reaction mixture was found to contain 0.304 moles of HI (g), 5.07×10-2 moles of H2 (g), and 4.57×10-2 moles of I2 (g), in a 1.00 liter container.Indicate True (T) or False (F) for each of the following:1. In order to reach equilibrium HI(g) must be produced .2. In order to reach equilibrium Kc must increase .3. In order to reach equilibrium H2 must be consumed .4. Qc is less than Kc.5. The reaction is at equilibrium. No further reaction will occur.

Answer 1

Answer:

Answer for the given statements: (1) T , (2) F , (3) T , (4) F , (5) F

Explanation:

At the given interval, concentration of HI = $\frac{0.304}{1}M=0.304M$

Concentration of $H_{2}$ = $\frac{5.07\times 10^{-2}}{1}M=5.07\times 10^{-2}M$

Concentration of $I_{2}$ = $\frac{4.57\times 10^{-2}}{1}M=4.57\times 10^{-2}M$

Reaction quotient,$Q_{c}$ , for this reaction = $\frac{[H_{2}][I_{2}]}{[HI]^{2}}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(5.07\times 10^{-2})\times (4.57\times 10^{-2})}{(0.304)^{2}}=2.51\times 10^{-2}$

So, the reaction is not at equilibrium.

As $Q_{c}> K_{c}$ therefore reaction must run in reverse direction to reduce $Q_{c}$ and make it equal to $K_{c}$. That means HI(g) must be produced and $H_{2}$ must be consumed.