For a system at equilibrium, the rate of the forward reaction equals the rate of

Question

For a system at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. For the reaction 3H_2 (g) + N_2(g) 2NH_3 (g) at equilibrium, a decrease in the pressure of the system would result in a shift toward the product. When Q is greater than K the reaction will form more products. If K_c for a reaction =5.0 times 10^4 K_c for the reverse reaction would equal 2.0 times 10^-5. A reaction with a K_c of 10^-2 is more reactant favored than a reaction with a K_c of 10^-3 With to the conversion of Kc to Kp, for the reaction: H_2 (g) + Br_2 (g) 2HBr (g) regard K_c would equal K_p. The calculated poH value of 5.0 M NaOH_(aq) solution is greater than 14. HPO_4^2- (aq) is the conjugate base of H_2PO_4^- (aq). Solutions of 0.20M HF (aq) and 0.020M HF (aq) would have the same % ionization. pK_w = pH + poH A solution with a pOH of 11 has a higher [H_3O^+] than a solution with a pH of 4. If water is added to a solution of 0.10 M HCl (aq), the numerical value of the pH will decrease.

Explanation / Answer

1. True

At equilibrium the rate of forward reaction equals to that of backward so the amount of reactants and products remain constant.

2. False

Since the pressure is already low in product side , so according to le chateliers principle , a decrease in pressure would favour shift towards reactants.

3. False

When Q (reaction quotient) is greater than equilibrium constant K , that means already a greater concentration of products is present in the system, hence the reaction favours the formation of reactants.

4. True

Kc for the reverse direction is given as kc' = 1/kc

So kc' = 1 / 5.0 × 104

kc' = 2 × 10-5

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