Question

4. What is the K for the reaction: 2Cu(s) + O2(g) 2CuO(s) 4Cu(s) +O2(g) 2Cu20(s) 4CUO(s) 2CU2O (s) + O2(g) K1 Given: K2 a) (Ki)(K2) b) (K2)2(Ki) c) (Ki)12(K2) d) (K2) (Ki)1 e) (Ki)(K2) -12

How do I do this? I know I'm supposed to manipulate the two equations but I have no idea how to combine the two equations to get the one from the question.

Answer 1

note that ,on reversing an equation k becomes 1/k

ki CS Required aquation - 20u + O 2 cuo Griven equations » 4 Cu + 2 (zo (s) 4 Cu O (s) - 20a₂O + O₂(g) Ke Now reverse ② equation then new k; long = Ke 2 Cu₃O7 02 4 Cu O - / -③ divide ③ - 2 then new k = S) new k= (1+2) 32 = (kz) / 2 & Cu₂O + 202 ²2 cuo - (K₂) 2 Now divide ① + 2 their new k= (K,)12 2Cu + 1/2O2 cuo - (k.)/ ③ Add 4 +6 at then new k= (K, J/(kg)? Cu2O 4112 2 +2eu & 140, 2040 + CuO (k, ) / (K) 12 2 cut U2 acuo (kil (k₂ )?

On dividing an equation by 2 ,k becomes (k)^1/2

On adding 2 equation , k becomes (k1 k2)

Thus answer is option C