PART I 1) Consider the following reaction: H2 (g) + Cl2 (g) 2 HCl (g) Keq-0.5 at
ID: 561020 • Letter: P
Question
PART I 1) Consider the following reaction: H2 (g) + Cl2 (g) 2 HCl (g) Keq-0.5 at 100 K What will be the Keq value if the above reaction is multiplied by three and the reaction is reversed. A) 1.5 B) 8 C) 16 D) 0.33 2) Consider the following reaction A(g) + 2B(g) 2C(g) + D(g) AH-250 kJ/mole When the temperature of the system is increased, what will happen A) The system will go towards the reactants side B) The volume of the system will decrease C) The reaction will go towards the product side D) No change will be seen Consider the following equilibrium at 200K The Keq of this reaction will change by 2 SOs(g) B) adding more SO2 (g) D) removing some 0-8) 3) 2 SO2 (g) + 02 (g) A) increasing pressure C) decreasing temperatureExplanation / Answer
1)
Answer
B) 8
Explanation
H2(g) + Cl2(g) <-------> 2HCl(g)
Keq = [HCl]^2/[H2] [Cl2] = 0.5
when it multiply by 3
3H2(g) + 3Cl2(g) <-------> 6HCl
Keq= [HCl]^6/[H2]^3[Cl2]^3
=( [HCl]^2/[H2][Cl2])^3
= (0.5)^3
= 0.125
when the reaction is reversed
Keq(rev) = 1/0.125 = 8
2)
Answer
A) The system will go towards reactants side
Explanation
This reaction is exothermic reaction , so when temperature is increased it will be used by reverse reaction which is endothermic
3)
Answer
C) Decreasing temperature
Explanation
Equillibrium constant is depands on temperature and not on other parameters
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.