Imibuzo Eyisibonelo kanye Nengxoxo Yokulingana Ezixazululweni
I-Pendahuluan
Ukulingana kwamakhemikhali kuwumqondo oyisisekelo kumakhemikhali obalulekile ekuqondeni ukusabela kwamakhemikhali. Ukulingana esixazululweni yisimo lapho ukusabela okubili noma ngaphezulu okuphambene kwenzeka khona ngesivinini esifanayo, ukuze amazinga ezinto ezisabelayo kanye nemikhiqizo ahlale engaguquki ngokuhamba kwesikhathi. Kulesi sihloko, sizoxoxa ngezibonelo eziningana zokulingana ezixazululweni futhi sixoxe ngazo ngokuningiliziwe. Sithemba ukuthi lesi sihloko sizokusiza ujulise ukuqonda kwakho inkolelo-mbono yokulingana kwamakhemikhali.
Imiqondo Eyisisekelo Yokulingana Kwamakhemikhali
Ngaphambi kokungena emibuzweni yesibonelo kanye nengxoxo, kubalulekile ukuqonda imiqondo eyisisekelo yokulingana kwamakhemikhali:
1. Ukulingana Okuguquguqukayo: Esimweni sokulingana, ukusabela akumi, kodwa ukusabela okuya phambili kanye nokusabela okuphambene kwenzeka ngesivinini esifanayo.
2. I-Equilibrium Constant (K): Isetshenziselwa ukukhombisa izinga lapho ukusabela kufinyelela khona ku-equilibrium. Ifomula ejwayelekile ye-equilibrium constant yile:
\[ K = \frac{{[product]}}{[reactant]} \]
3. Umthetho Wesenzo Esiningi: Ukuhlushwa kwento ngayinye ekuphenduleni kuvezwa ngesimo se-exponential ngokuya ngama-coefficients ku-chemical equation.
Imibuzo Eyisibonelo Nengxoxo
Isibonelo Umbuzo 1
Umbuzo:
I-equation yokusabela kwe-equilibrium yaziwa kanje:
\[ N_2O_4 (g) \umcibisholo wesokunxele 2NO_2 (g) \]
Ekushiseni okuthile, ukuhlushwa kwe-equilibrium kwe-\( N_2O_4 \) kungu-0.10 M kanye ne-\( NO_2 \) kungu-0.20 M. Bala inani le-equilibrium constant, \( K_c \), ukuze uthole impendulo.
Ingxoxo:
Isibalo se-equilibrium constant \( K_c \) sithi:
\[ K_c = \frac{{[NO_2]^2}}{[N_2O_4]} \]
Faka amanani okuhlushwa ku-equation:
\[ K_c = \frac{{(0.20)^2}}{0.10} \]
\[ K_c = \frac{{0.04}}{0.10} = 0.40 \]
Ngakho-ke, inani le-\( K_c \) lokusabela lingu-0.40.
Isibonelo Umbuzo 2
Umbuzo:
Esitsheni esivaliwe, igesi ye-\( PCl_5 \) ibola ibe yi-\( PCl_3 \) kanye ne-\( Cl_2 \) ngokwe-equation:
\[ PCl_5 (g) \leftrightarrow PCl_3 (g) + Cl_2 (g) \]
Uma i-equilibrium constant, \( K_c \), ekushiseni okuthile ingu-0.200 futhi ekushiseni okulinganayo i-concentration ye-\( PCl_5 \) ingu-1.00 M, bala i-concentrations ye-\( PCl_3 \) kanye ne-\( Cl_2 \) ekushiseni okulinganayo.
Ingxoxo:
Ake sithi ukuhlushwa kwe-\( PCl_3 \) kanye ne-\( Cl_2 \) ekulinganisweni kungu-x M. Ngokusekelwe ku-equation yokusabela, ukuhlushwa kwe-\( PCl_5 \) kwehla ngo-x M futhi.
Ngakho-ke, i-equation yokulingana ingabhalwa kanje:
\[ K_c = \frac{{[PCl_3][Cl_2]}}{[PCl_5]} \]
Faka amanani e-\( K_c \) kanye nokuhlushwa kwe-\( PCl_5 \):
\[ 0.200 = \frac{{x^2}}{(1.00 – x)} \]
Ukuxazulula izilinganiso ze-quadratic:
\[ 0.200 = \frac{{x^2}}{(1.00 – x)} \]
\[ 0.200 (1.00 – x) = x^2 \]
\[ 0.200 – 0.200x = x^2 \]
\[ x^2 + 0.200x – 0.200 = 0 \]
Isixazululo salesi sibalo se-quadratic ngu-x = 0.332 noma u-x = -0.532 (isixazululo esingesihle asibalulekile, njengoba ukuhlushwa kungeke kube kubi).
Ngakho-ke, ukuhlushwa kwe-\( PCl_3 \) kanye ne-\( Cl_2 \) ekulinganisweni kungu-0.332 M.
Isibonelo Umbuzo 3
Umbuzo:
Esixazululweni, ama-ion e-hydroxide (\( OH^- \)) angasabela nama-ion e-ammonium (\( NH_4^+ \)) ukuze akhe amanzi (\( H_2O \)) kanye ne-ammonia (\( NH_3 \)). I-equation yokusabela yile:
\[ NH_4^+ (aq) + OH^- (aq) \leftrightarrow H_2O (l) + NH_3 (aq) \]
Uma i-equilibrium constant (\( K_c \)) isethwe ku-1.8 x 10^5 ekushiseni okuthile, futhi amazinga okuqala e-\( NH_4^+ \) kanye ne-\( OH^- \) angu-0.10 M kanye no-0.15 M, ngokulandelana, bala amazinga e-\( NH_3 \) lapho amazinga e-equilibrium efinyelelwe.
Ingxoxo:
Bhala i-equation yokulingana:
\[ K_c = \frac{{[NH_3]}}{[NH_4^+][OH^-]} \]
Ake sithi u-x uwukuhlushwa kuka-\( NH_3 \) okwakhiwe ngesikhathi sokulingana, khona-ke ukuhlushwa kuka-\( NH_4^+ \) kanye no-\( OH^- \) kwehla ngo-x ngokulandelana.
\[ K_c = \frac{{x}}{(0.10 – x)(0.15 – x)} \]
Faka inani le-\( K_c \):
\[ 1.8 \izikhathi ezingu-10^5 = \frac{{x}}{(0.10 – x)(0.15 – x)} \]
Ake sithi u-x mncane kakhulu uma uqhathaniswa no-0.10 no-0.15 kangangokuthi \( 0.10 – x \cishe 0.10 \) kanye \( 0.15 – x \cishe 0.15 \):
\[ 1.8 \izikhathi ezingu-10^5 = \frac{{x}}{(0.10)(0.15)} \]
\[ x = 1.8 \izikhathi ezingu-10^5 \izikhathi (0.10) \izikhathi (0.15) \]
\[x = 2.7 \]
Ngakho-ke, ukuhlushwa kwe-\( NH_3 \) ekulinganisweni kulinganiselwa ku-2.7 M.
Isiphetho
Kusukela engxoxweni engenhla, kungaphethwa ngokuthi ukuqonda umqondo wokulingana kwamakhemikhali kubalulekile ekuxazululeni izinkinga ezahlukahlukene ezihlobene nokulingana ezixazululweni. Sisebenzisa umthetho wesenzo esikhulu kanye ne-equilibrium constant, singabala amazinga ezinhlobo ezahlukene ekuphenduleni lapho kufinyelelwa ukulingana. Lesi sihloko sithinta kuphela izibonelo ezimbalwa eziyisisekelo. Empeleni, ukubalwa kokulingana kwamakhemikhali kungaba yinkimbinkimbi kakhulu, okubandakanya izinto ezifana nokushisa, ingcindezi, kanye nomsebenzi we-ion esixazululweni.
