Answer to Question #152383 in General Chemistry for Michael Gordon

Question

The electrolysis of 1L-aqueous solution of copper (II) nitrate is taking place. The initial pH at the cathode is 1.00. The electrolysis is carried out using a current of 2.45 A and when the electrolysis is over, the pH at the cathode is 1.05.
Determine the reaction time.
When the gases formed at the anode and cathode are mixed, the following equilibrium takes place:
2 NO(g) + O2(g) ⇌ 2 NO2(g)
The concentration of NO2 once the equilibrium is reached is 2.51 x 10–3 mol/L. Determine the equilibrium constant.

Answer 1

$2NO_{(g)} +O_{2(g)} \to 2NO_{2(g)}$

With 2moles of NO and 1 mole of O2 in 1litre flask.

2.5 X 10-3[NO] + 1.25X 10-3[O2] -> 2.51 X 10-3[NO2] at equilibrium.

Hence equilibrium constant,

Kc = $\dfrac{[NO_2]^2}{[NO]^2[O2]}$ = $\dfrac{[2.51×10^{-3}]^2}{ [2.51 ×10^{-3}]^2 ×[ 1.25 × 10^{-3}]}$ = 8×10²

Since Kc >> 1 reaction is almost complete and equilibrium exists at the product.