化学平衡知识点归纳 与经典习题赏析
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Transcript of 化学平衡知识点归纳 与经典习题赏析
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(1)(2)
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1 aA(g) bB(g) + cC(g)AA abc A abc A abc A
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2: aA(g) + bB(g) cC(g) + dD (g) AAB AB abcd AB abcdAB abcdAB
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(1) (2)aA(g)+bB(g) cC(g)cmol C(g) amolA (g)bmolB(g) x molA(g )ymolB(g) zmol C(g) (x+az/c)molA(g)(y+bz/c)molB(g)
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ABSO2C
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COABCOCOCD
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3. 201214T2.0 L1.0 mol PCl5PCl5(g) PCl3(g) + Cl2(g) A.50 s v(PCl3)0. 0032 molL1s1 B.c(PCl3)0.11 molL1H0 C.1.0molPCl50. 20 molPCl30. 20mol Cl2v()v() D.2. 0 mol PCl3 2. 0 mol Cl2PCl3 80%
t / s050150250350n(PCl3) / mol00.160.190.200.20
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Av(PCl3)0.16mol2L50s 0.0016molL1s1 Bc(PCl3)0.1mol/LPCl3 CK0.10.10.40.025Qc0.10.10.50.02QKv()v() D1mol PCl51mol PCl31mol Cl2PCl3(10.2)1100802mol PCl32mol Cl2PCl380
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4. 20126 2SO2(g) + O2(g) 2SO3(g) H197 kJmol1() 2 mol SO21 mol O2() 1 mol SO20.5 mol O2() 2 mol SO3 A. PPP2P B. SO3mmm2m C. c(SO2)c(O2)kkkk D. Q QQ2Q
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( )
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ppSO3mm1/21/2SO32ppSO32mmABSO2O22121SO2O221SO2O221CSO3SO3D
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5. 201215A(g) + B(g) C(g) + D(g)
1K H 01Kc(C)c(D)/c(A)c(B) KH0
700900830100012001.71.11.00.60.4
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28305 L0.20molA0.80molB6sA(A)0.003 molL1s16sc(A)_____molL1 C_______molA________1 molA______________2(A)0.003 molL1s1 Ac(A)(A)t0.018 molL1A0.2mol5L0.018molL10.022molL1A0.018 molL15L0.09molC0.09mol0.0220.0980%80%
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830Ax A(g) + B (g)C(g) + D(g) (molL1) 0.04 0.16 0 0(molL1) x x x x(molL1) 0.04x 0.16x x x
xx/(0.04x)(0.16x)1 x0.032A:(A)0.032mol/L0.04mol/L100%80%
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3 () a. b. c. c(A) d. CD (3)abc(A)cCDdc
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41200C(g) + D(g) A(g) + B(g) (1)H< 0(2)c(A)0.20 mol/5 L0.04 molL1v(A)c(A)0.018 molL1c(A)c(A)c(A)0.022 molL1(3)abd(4)2.5
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(2011)X(g) + 2Y(g) 2Z(g) 2M(g) N(g) + P(g)
A. B. I14:15C. IX5/11D. IIIM
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ABPV=nRTTVnI PV=nRT nRTP1V1=P2V2 P2/P1=V1/V2= 2/2.2=10/11C T PV=nRT PV/n=RT P1V1/n1=P2V2/n2 n2=P2V2n1/P1V1= P2/P1V2/V1n1=10/112.8/33=28/1128/11IXYZ(1m)(22m)2m28/11m=5/11CDIIIM
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1N2H2NH36:8:1N2H2NH39:27:8 A. 75% B. 50% C. 25% D. 20%
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N2:H2:NH3 6:8:1N2:H2:NH3=9:27:8N29molnN=92+8=26molN26molnN=62+1=13mol NN212mol(122+2=26)12mol9mol3mol N23mol
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2ABA(g)+2B(g) 2C(g)ABC A.40% B.50% C.60% D.70%
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ABAAB1:2ABABA50%A50%AA
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33molX2molY 4X(g)+4Y(g) 3Q(g)+nR(g)10%X1/3n A.3 B.4 C.5 D.7
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4X(g)+4Y(g) 3Q(g)+nR(g) 3 2 0 0 3 =1 1 2 1 0.75n=2+1+0.75+ 3.75+ 3.75+ 5.5 n=7
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4X(g)+4Y(g) 3Q(g)+nR(g) 4 n+3(4+4) 3 (3+2)10%4(3+2)10%3 (n5) n7
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X4+43+nn5D