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Dalam ruangan yang tekanannya 3 atm, dipanaskan 0,5 mol gas  dan 1,5 mol gas . Pada suhu 400 K, terjadi reaksi kesetimbangan: Ternyata pada saat setimbang terdapat gas  sebanyak 0,25 mol. Hitunglah  dan  pada saat itu.

Pertanyaan

Dalam ruangan yang tekanannya 3 atm, dipanaskan 0,5 mol gas begin mathsize 14px style N subscript 2 end style dan 1,5 mol gas begin mathsize 14px style H subscript 2 end style. Pada suhu 400 K, terjadi reaksi kesetimbangan:


begin mathsize 14px style N subscript 2 open parentheses italic g close parentheses space plus space 3 H subscript 2 open parentheses italic g close parentheses space rightwards harpoon over leftwards harpoon space 2 N H subscript 3 open parentheses italic g close parentheses end style


Ternyata pada saat setimbang terdapat gas undefined sebanyak 0,25 mol. Hitunglah begin mathsize 14px style K subscript p end style dan begin mathsize 14px style K subscript c end style pada saat itu.

  1.  ...undefined 

  2.  ...undefined 

Pembahasan Soal:

1. Buat sistem MRS untuk mencari jumlah mol pada saat setimbang.

 

2. Hitung tekanan parsial masing-masing zat pada keadaan setimbang.

begin mathsize 14px style P subscript N subscript 2 end subscript equals n over n subscript total cross times P subscript total P subscript N subscript 2 end subscript equals fraction numerator 0 comma 25 space mol over denominator 1 comma 5 space mol end fraction cross times 3 space atm P subscript N subscript 2 end subscript equals 0 comma 5 space atm  P subscript H subscript 2 end subscript equals n over n subscript total cross times P subscript total P subscript H subscript 2 end subscript equals fraction numerator 0 comma 75 space mol over denominator 1 comma 5 space mol end fraction cross times 3 space atm P subscript N subscript 2 end subscript equals 1 comma 5 space atm  P subscript N H subscript 3 end subscript equals n over n subscript total cross times P subscript total P subscript N H subscript 3 end subscript equals fraction numerator 0 comma 5 space mol over denominator 1 comma 5 space mol end fraction cross times 3 space atm P subscript N H subscript 3 end subscript equals 1 comma 0 space atm end style 

3. Hitung begin mathsize 14px style K subscript p end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript p end cell equals cell fraction numerator open parentheses P subscript N H subscript 3 end subscript close parentheses squared over denominator open parentheses P subscript N subscript 2 end subscript close parentheses open parentheses P subscript H subscript 2 end subscript close parentheses cubed end fraction end cell row cell K subscript p end cell equals cell fraction numerator blank over denominator left parenthesis 0 comma 5 right parenthesis left parenthesis 1 comma 5 right parenthesis cubed end fraction end cell row cell K subscript p end cell equals cell 0 comma 59 end cell end table end style      

4. Hitung nilai begin mathsize 14px style K subscript c end style dengan menggunakan rumus hubungan undefined dan undefined.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row T equals cell 400 space K end cell row R equals cell 0 comma 082 space L point atm space mol to the power of negative sign 1 end exponent K to the power of negative sign 1 end exponent end cell row cell triangle n end cell equals cell koefisien space produk bond koefisien space reaktan end cell row cell triangle n end cell equals cell 2 minus sign 4 equals minus sign 2 end cell row blank blank blank row cell K subscript p end cell equals cell K subscript c open parentheses R point T close parentheses to the power of increment n end exponent end cell row cell 0 comma 89 end cell equals cell K subscript c left parenthesis 0 comma 082.400 right parenthesis to the power of negative sign 2 end exponent end cell row cell K subscript c end cell equals cell fraction numerator 0 comma 59 over denominator left parenthesis 0 comma 082.400 right parenthesis to the power of negative sign 2 end exponent end fraction end cell row cell K blank subscript c end cell equals cell fraction numerator 0 comma 59 over denominator begin display style bevelled 1 over left parenthesis 0 comma 082.400 right parenthesis squared end style end fraction end cell row cell K blank subscript c end cell equals cell 0 comma 59 cross times left parenthesis 0 comma 082.400 right parenthesis squared over 1 end cell row cell K blank subscript c end cell equals cell 634 comma 75 end cell row blank blank blank end table end style        

Jadi, nilai begin mathsize 14px style K subscript p space end subscript end styledan undefined berturut-turut adalah 0,59 dan 634,75.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 05 Maret 2021

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