I need help with equilibrium. I\'m not sure where to start for these. Equilibriu

Question

I need help with equilibrium. I'm not sure where to start for these. Equilibrium Worksheet Write the Q expression for the following reactions: 2. 2NOtg) &H--569; kunnol. For the reaction N040 What happens to Q and K, for each of the following? Which way will the reaction shift? a) remove N.O b) raise the temperature For the following reactions, what happens if the volume of the container is decreased Hint: the size of the container only affects the pressure and concentration of a gas 3. a) A(aq) + B(s) C() + D(g) c) 2G(g) H(g) Initially, 0.40 moles of PCl, are put in a 10.0 L container. At equilibrium, 0.25 moles of Cl are present. What is the value of K,? Hint: you can fill in the whole table without any x's 5. H2(g) + 12(g) 2H1(g) The equilibrium concentrations are [H:]-0.46 M, [lJ-0.39 M, and [HI 3 M. Calculate K

Explanation / Answer

1)The reaction quotient or concentration quotient Q for any general reaction aA+bB<----> cC+dD is given as,

Q = [C]c[D]d / [A]a[B]b ..........(I) (see that coefficients are raised as powers of concentration of the species.)

Hence for the given reactions we can write,

a) 2H2(g) + O2(g) -----------> 2H2O(g).

Q = [H2O]2 / [H2]2[O2].

b) NH3 (aq.) + H2O (l) <---------> NH4+(aq.)+ OH- (aq.) .................(Homogeneous reaction in liquid phase)

Q = [OH-] [NH4+] / [H2O] [NH3].

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2) N2O4(g)<-----> 2NO2(g) Delta H = -56.9kJ/mol.

Q and Kc are same at equilibrium stage and we can write Q=Kc = [NO2]2 / [N2O4].

Various factors have effect on the equilibrium state. Changes in concentration of reactant and product, temperature , pressure favours either forward or backward reaction. This is stated as Le-Chatelier's priniciple, that every costrain impossed on the state in equilibrium, the system respond it in a way that the effect should get nulified and equilibrium remains constant.

Hence, i) Removal of the reactant N2O4 lowers the concentration of N2O4 i.e. reactant and hence the value of Q and Kc will increase.

ii) The enthaly change (Delta H) is negative. This means the reaction is exothermic. I.e. heat is released to the surrounding. Hence increase in temperature disfavours forward reaction and the backward reaction i.e. formation of N2O4 is favoured. This increases the concentration of N2O4 and will lower that of NO2.Hence Q or Kc value lowers.

We can say that, for exothermic reaction the increase in temperature equilibrium constant Kc or Q decreases.

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