《无机化学》课程教学资源（PPT课件）第三章 化学反应速率

热力学解决了化学反应的可能性问题，自发过程是否一定进行得很快?△,G° = -4.78 kJ·mol -12NO,(g) = N,O4 (g)实际上，反应速率相当快！H2(g) +=O2(g) =H,O()△,G = -237.18 kJ·mol-1l2实际上，反应速率相当慢！热力学可能性动力学---现实性

热力学解决了化学反应的可能性问题, 自 发过程是否一定进行得很快? 实际上，反应速率相当慢! 实际上, 反应速率相当快! 热力学- 可能性 动力学- 现实性 θ 1 2NO2 (g) N2 O4 (g) Δr m 4 78 kJ mol − = G = − . • θ 1 2 O2 (g) H2 O(l) Δr m 237 18 kJ mol 2 1 H (g) − + = G = − . •

Figure. Reaction rates. The rates of chemical reactions span arange of time scales. For example, explosions are rapid,occurring in seconds or fractions of seconds; corrosion cantake years; and the weathering of rocks takes place overthousands or even millions of years.这些都是化学动力学的研究范畴

这些都是化学动力学的研究范畴。 Figure. Reaction rates. The rates of chemical reactions span a range of time scales. For example, explosions are rapid, occurring in seconds or fractions of seconds; corrosion can take years; and the weathering of rocks takes place over thousands or even millions of years

第3章化学反应速率

第3章 化学反应速率

3-1反应速率的定义平均速率3-1-1单位时间内浓度的改变量为基础来研究的单位：mol : dm-3. s-1mol : dm-3 : min-1mol : dm-3 : h-1平均速率：指某一时间间隔内的化学反应速率的平均值

3-1 反应速率的定义 单位时间内浓度的改变量为基础来研究的 mol ·dm-3 · s -1 mol ·dm-3 ·min-1 mol ·dm-3 ·h -1 3-1-1 平均速率 平均速率：指某一时间间隔内的化学反应 速率的平均值。 单位:

例如，340 K时N,O.的热分解反应：2N20s 4NO2+O2平均速率二Ac(N,O,)正值r(N,Os)=△t△c(NO2)正值r(NO2) =△t△c(O2)正值r(O2) =△t

平均速率 例如，340 K 时N2O5的热分解反应： 正值 (N O ) (N O ) 2 5 2 5 t c r  −  = 正值 (N O ) (N O ) 2 2 t c r   = 2N2O5 4NO2+O2 正值 (O ) (O ) 2 2 t c r   =

2N2054NO2+O2 (T = 340:K)时间c(N2O5)r(min)(mol-dm-3-min-1)(mol-dm-3)00.0470.16010.1130.0B320.080.0.02430.0160.05640.040

时间 c(N2O5 ) (min) (mol·dm–3 ) (mol·dm–3·min–1 ) 0 0.160 0.047 1 0.113 0.033 2 0.080 0.024 3 0.056 0.016 4 0.040 2N2O5 4NO2+O2 r (T = 340 K)

2N,054NO2+O2当△t=4min时，测得上述反应的平均反应速率(mol·dm-3·min-l)为(N,0,)=c(N,}= 0.03 mol dm min -△tAc(NO,) = 0.06 mol · dm-3 ·min r(NO2) =△t△c(O2)) = 0.015 mol .dm-3 .minr(O2) =△t17r(N,Os)=二r(NO2)=r(O2)-1241

(O ) 1 1 (N O ) 4 1 (N O ) 2 1 2 5 2 2 r = r = r 2N2O5 4NO2+O2 2 5 3 1 2 5 0 03mol dm min N O N O − − = • •  −  = . t c( ) r( ) 当t = 4 min 时，测得上述反应的平均反 应速率(mol·dm–3·min–1 ) 为 2 3 1 2 0.06 mol dm min (N O ) (N O ) − − = • •   = t c r 2 3 1 2 0.015 mol dm min (O ) (O ) − − = • •   = t c r

aA+bB=gG+hH1 △c(A)1 △c(B)1 △c(G)1 △c(H)bh△t△t△t△tag1-(A)==r(B) =二r(G)=r(H)bhag

a A+ b B g G + h H t c t h c t g c t b c a   =   =   = −   − 1 (A) 1 (B) 1 (G) 1 (H) (H) 1 (G) 1 (B) 1 (A) 1 r h r g r b r a = = =

瞬时速率3-1-2c(NO)FigureA graph of the concentration of NOr = 7.7 X 10-5 mol·dm-3.s-1againsttimeinthereactionNO,+CONO+CO,.The average rateover0.0326thetimeintervalfrom50to15000300seconds is obtained by dividing the0.0288change in NO concentration by thedurationof the interval (green box)0.0249The instantaneous rate 150 secondsafterthe startofthereactionisfound0.0200by calculating the slope of the linetangent to the curve at that point (red0.0160box).瞬时速率是△tAveragerate0.0100趋近于0时，平=1.3 X 10-4均速度的极限。mol·dm-3.s-1200r(NO)=dy/dx= dc(NO)/dt (切线的斜率)9

9 瞬时速率是 t 趋近于0时，平 均速度的极限。 r(NO) =dy/dx= dc(NO)/dt (切线的斜率) c(NO) r = 7.7 ×10-5 mol·dm-3·s-1 Average rate =1.3×10-4 mol·dm-3·s-1 3-1-2 瞬时速率 Figure A graph of the concentration of NO against time in the reaction NO2 + CO → NO + CO2 . The average rate over the time interval from 50 to 150 seconds is obtained by dividing the change in NO concentration by the duration of the interval (green box). The instantaneous rate 150 seconds after the start of the reaction is found by calculating the slope of the line tangent to the curve at that point (red box)

NO + CO2NO2 + CO△c(NO)dc(NO)r(NO) = limdt△t△t>0

t c t c r t d (NO) d (NO) (NO) lim 0 =   =  → NO2 + CO NO + CO2