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Perhatikan proses kesetimbangan pada  : Konse...

Perhatikan proses kesetimbangan pada begin mathsize 14px style 686 space degree C end style :


begin mathsize 14px style C O subscript 2 open parentheses italic g close parentheses space plus space H subscript 2 open parentheses italic g close parentheses space rightwards harpoon over leftwards harpoon space C O open parentheses italic g close parentheses space plus space H subscript 2 O open parentheses italic g close parentheses end style


Konsentrasi kesetimbangan spesi-spesi yang bereaksi adalah begin mathsize 14px style open square brackets C O close square brackets equals 0 comma 050 space M end stylebegin mathsize 14px style open square brackets H subscript 2 close square brackets equals 0 comma 045 space M end stylebegin mathsize 14px style open square brackets C O subscript 2 close square brackets equals 0 comma 086 space M end style, dan begin mathsize 14px style open square brackets H subscript 2 O close square brackets equals 0 comma 040 space M end style.

(a) Hitunglah begin mathsize 14px style K subscript italic c end style untuk reaksi pada begin mathsize 14px style 686 space degree C end style.

(b) Jika konsentrasi begin mathsize 14px style C O subscript 2 end style ditingkatkan menjadi 0,50 mol/L dengan menambahkan begin mathsize 14px style C O subscript 2 end style, berapa konsentrasi semua gas ketika kesetimbangan tercapai kembali?undefined

Jawaban:

Tetapan kesetimbangan (begin mathsize 14px style K subscript italic c end style) merupakan perbandingan hasil kali konsentrasi produk dan reaktan dipangkatkan koefisien reaksinya masing-masing. Untuk menghitung nilai begin mathsize 14px style K subscript italic c end style, konsentrasi zat yang dimasukkan kerumus begin mathsize 14px style K subscript italic c end style hanya zat yang fasenya gas (g) dan larutan (aq) saja. Harga begin mathsize 14px style K subscript italic c end style dipengaruhi oleh suhu, selama suhu tetap maka harga begin mathsize 14px style K subscript italic c end style akan tetap.

Dengan demikian, perhitungannya adalah sebagai berikut.

a. Nilai begin mathsize 14px style K subscript italic c end style untuk reaksi pada begin mathsize 14px style 686 space degree C end style yaitu :


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript italic c end cell equals cell fraction numerator open square brackets C O close square brackets. open square brackets H subscript 2 O close square brackets over denominator open square brackets C O subscript 2 close square brackets. open square brackets H subscript 2 close square brackets end fraction end cell row blank equals cell fraction numerator left parenthesis 0 comma 05 right parenthesis. left parenthesis 0 comma 04 right parenthesis over denominator left parenthesis 0 comma 086 right parenthesis. left parenthesis 0 comma 045 right parenthesis end fraction end cell row blank equals cell fraction numerator 0 comma 002 over denominator 0 comma 00387 end fraction end cell row blank equals cell 0 comma 52 end cell end table end style


b. Jika konsentrasi begin mathsize 14px style C O subscript 2 end style ditingkatkan menjadi 0,50 mol/L maka :



Nilai x dapat dihitung dengan rumus berikut.


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript italic c end cell equals cell fraction numerator open square brackets C O close square brackets. open square brackets H subscript 2 O close square brackets over denominator open square brackets C O subscript 2 close square brackets. open square brackets H subscript 2 close square brackets end fraction end cell row cell 0 comma 52 end cell equals cell fraction numerator left parenthesis 0 comma 086 plus x right parenthesis. left parenthesis 0 comma 04 plus x right parenthesis over denominator left parenthesis 0 comma 5 minus sign x right parenthesis. left parenthesis 0 comma 045 minus sign x right parenthesis end fraction end cell row cell 0 comma 52 end cell equals cell fraction numerator 0 comma 0034 plus 0 comma 126 x plus x squared over denominator 0 comma 0225 minus sign 0 comma 545 x plus x squared end fraction end cell row cell 0 comma 0117 minus sign 0 comma 2834 x plus 0 comma 52 x squared end cell equals cell 0 comma 0034 plus 0 comma 126 x plus x squared end cell row cell left parenthesis 0 comma 0117 minus sign 0 comma 2834 x plus 0 comma 52 x squared right parenthesis minus sign left parenthesis 0 comma 0034 plus 0 comma 126 x plus x squared right parenthesis end cell equals 0 row cell 0 comma 0083 minus sign 0 comma 1574 x minus sign 0 comma 48 x squared end cell equals 0 row a equals cell negative sign 0 comma 48 semicolon space b equals minus sign 0 comma 1547 semicolon space c equals 0 comma 0083 end cell row blank blank blank row blank blank cell difaktorkan comma space maka space colon end cell row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative sign b space plus-or-minus space square root of b squared minus sign 4 ac end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator 0 comma 1547 plus-or-minus square root of left parenthesis minus sign 0 comma 1547 right parenthesis squared minus sign 4 left parenthesis minus sign 0 comma 48 right parenthesis left parenthesis 0 comma 0083 right parenthesis end root over denominator 2 left parenthesis minus sign 0 comma 48 right parenthesis end fraction end cell row blank equals cell fraction numerator 0 comma 1547 plus-or-minus square root of 0 comma 04 end root over denominator negative sign 0 comma 96 end fraction end cell row blank equals cell fraction numerator 0 comma 1547 plus-or-minus 0 comma 2 over denominator negative sign 0 comma 96 end fraction end cell row cell x subscript 1 end cell equals cell fraction numerator 0 comma 1547 plus 0 comma 2 over denominator negative sign 0 comma 96 end fraction equals minus sign 0 comma 37 end cell row blank blank atau row cell x subscript 2 end cell equals cell fraction numerator 0 comma 1547 minus sign 0 comma 2 over denominator negative sign 0 comma 96 end fraction equals 0 comma 047 end cell end table end style


Nilai x yang dipakai adalah yang positif yakni 0,047 sehingga, konsentrasi masing-masing senyawa setelah kesetimbangan yaitu :


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets C O subscript 2 close square brackets end cell equals cell 0 comma 5 minus sign x end cell row blank equals cell 0 comma 5 minus sign 0 comma 047 end cell row blank equals cell 0 comma 453 end cell row cell open square brackets H subscript 2 close square brackets end cell equals cell 0 comma 045 minus sign x end cell row blank equals cell 0 comma 045 minus sign 0 comma 047 end cell row blank equals cell negative sign 0 comma 02 end cell row cell open square brackets C O close square brackets end cell equals cell 0 comma 086 plus x end cell row blank equals cell 0 comma 086 plus 0 comma 047 end cell row blank equals cell 0 comma 133 end cell row cell open square brackets H subscript 2 O close square brackets end cell equals cell 0 comma 04 plus x end cell row blank equals cell 0 comma 04 plus 0 comma 047 end cell row blank equals cell 0 comma 087 end cell end table end style


Jadi, berdasarkan perhitungan diatas sepertinya terdapat kesalahan pada soal karena tidak mungkin konsentrasi senyawa bernilai minus (konsentrasi begin mathsize 14px style H subscript 2 end style= -0,02 M).undefined

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