Answer:
Answer for the given statements: (1) T , (2) F , (3) T , (4) F , (5) F
Explanation:
At the given interval, concentration of HI = [tex]\frac{0.304}{1}M=0.304M[/tex]
Concentration of [tex]H_{2}[/tex] = [tex]\frac{5.07\times 10^{-2}}{1}M=5.07\times 10^{-2}M[/tex]
Concentration of [tex]I_{2}[/tex] = [tex]\frac{4.57\times 10^{-2}}{1}M=4.57\times 10^{-2}M[/tex]
Reaction quotient,[tex]Q_{c}[/tex] , for this reaction = [tex]\frac{[H_{2}][I_{2}]}{[HI]^{2}}[/tex]
species inside third bracket represents concentrations at the given interval.
So, [tex]Q_{c}=\frac{(5.07\times 10^{-2})\times (4.57\times 10^{-2})}{(0.304)^{2}}=2.51\times 10^{-2}[/tex]
So, the reaction is not at equilibrium.
As [tex]Q_{c}> K_{c}[/tex] therefore reaction must run in reverse direction to reduce [tex]Q_{c}[/tex] and make it equal to [tex]K_{c}[/tex]. That means HI(g) must be produced and [tex]H_{2}[/tex] must be consumed.
