Answer:
a) 1,5x10³ā“
b) 1,5x10¹āµ
c) 7,3x10¹ā“
Explanation:
It is possible to sum different reactions with its equilibrium constant to obtain the equilibrium constant of the new reaction, thus:
a) 2 NO(g) ā Nā(g) + Oā(g) kc = 1/4,3x10ā»Ā²āµ
+ 2 NO(g) + Oā(g) ā 2 NOā(g) kc = 6,4x10ā¹
= 4 NO(g) ā Nā(g) + 2NOā(g) Where its equilibirum constant is:
1/4,3x10ā»Ā²āµ Ć 6,4x10ā¹ = 1,5x10³ā“
b) 4 NO(g) ā 2 Nā(g) + 2 Oā(g) kc = 2Ć 1/4,3x10ā»Ā²āµ
4 NOā(g) Ā ā 4 NO(g) + 2 Oā(g) kc = 2Ć 1/6,4x10ā¹
= 4 NOā(g) ā 2 Nā(g) + 4 Oā(g) Where its equilibirum constant is:
2Ć 1/4,3x10ā»Ā²āµ Ć 2Ć 1/6,4x10ā¹ = 1,5x10¹āµ
c) 4 NO(g) ā 2 Nā(g) + 2 Oā(g) kc = 2Ć 1/4,3x10ā»Ā²āµ
2 NOā(g) Ā ā 2 NO(g) + Oā(g) kc = Ā 1/6,4x10ā¹
= 2NO(g) + 2 NOā(g) ā 2 Nā(g) + 3 Oā(g) Where its equilibirum constant is:
2Ć 1/4,3x10ā»Ā²āµ Ć 1/6,4x10ā¹ = 7,3x10¹ā“
I hope it helps!