Question

(d) Kz/K
x 10-18
XeO2(g) + 2HF(g) XeO3F2(g) + H2O(g)
(a) KK2
(b) K/K2
7. Given the equilibrium constant values:
(1) N2(g) + O2(8) N2O(g); Kc=2.7 x 10-18
(II) N204(9) 2N2O(g); Kc = 4.6 x 10-3
(III) – N2 (9) + O2(9) — NO2(8); Kc=4.1 10-9
Thus, for the reaction, 2N2O(g) + 302(g) 2N,04(E),
Kis
(a) 5.46 x 10^7
(b) 5.46 x 10^-7
d) 1.832 x 10^-6
(C) 1.832 x 10^6​

Answer 1

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