Consider the kJ, and K_p = 9.3 times 10^-8 at 700 K: 2 H_2S(g) 2 H_2(g) + s_2(g)

Question

Consider the kJ, and K_p = 9.3 times 10^-8 at 700 K: 2 H_2S(g) 2 H_2(g) + s_2(g) Which way will the equilibrium be shifted (right, left, or not shifted) if the temperature is raised? the volume of the vessel is increased? the partial pressure of S_2 is increased? a catalyst is added? What is the value of Delta n(g) for the above equilibrium? Does the above equilibrium at 700 K favor reactants, products, or neither reactants or products? What is the conjugate base of H_2O ? H_3O OH+ O^2- H_2O^+ OH^-

Explanation / Answer

3. a) Back ward reaction ( Equilibrium shift Left direction ) takes place

b) volume does not influences the equilibrium

c) Back ward reaction ( Equilibrium shift Left direction ) takes place

d) The net effect of catalyst is zero

4. Delta n = 3-2 =1 ( sum of moles of gaseous products - sum of moles of gaseous reactants)

5. at 700k , the Back ward reaction ( Equilibrium shift Left direction ) takes place , means it is favour of reactants

6. OH- - conjugate base of H2O - ANS - E

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