Chapter%209

Chapter%209 - 9.8 9.20 (a) K = PNO2 2 2 PNO PO2 ; (b) K =...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
9.8 (a) 2 2 2 NO 2 NO O = P K P P ; (b) 3 2 5 SbCl Cl SbCl = P P K P ; (c) 2 4 2 2 N H 2 N H = P K P P 9.20 r f 2 1 (CO , g) 394.36 kJ mol Δ ° = Δ ° = − G G 3 1 1 1 69 ln ln 394.36 10 J mol ln 159.17 (8.314 J K mol )(298 K) 1.3 10 Δ ° = − Δ ° = − × = − = + = × G RT K G K RT K K In practice, no K will be so precise. A better estimate would be 69 1 10 × . Because Q < K , the reaction will tend to proceed to produce products. 9.32 H 2 (g) + Cl 2 (g) 2 HCl(g) 8 5.1 10 = × C K 2 2 2 [HCl] [H ][Cl ] = C K 3 2 8 3 2 3 2 2 8 3 12 1 2 (1.45 10 ) 5.1 10 [H ](2.45 10 ) (1.45 10 ) [H ] (5.1 10 )(2.45 10 ) [H ] 1.7 10 mol L × × = × × = × × = × 9.36 (a) 2 3 3 2 2 [NH ] 0.278 [N ][H ] = = C K 2 3 [0.122] 0.248 [0.417][0.524] = = C Q (b) C C Q K ; therefore the system is not at equilibrium. (c) Because < C C Q K , more products will be formed.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
9.58
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

Chapter%209 - 9.8 9.20 (a) K = PNO2 2 2 PNO PO2 ; (b) K =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online