1. Sekolah Dasar
  2. Sekolah Menengah Pertama
  3. Sekolah Menengah Atas
  4. Mata Pelajaran
  5. Programming

Mohon bantuannya untuk soal ini​

Berikut ini adalah pertanyaan dari Titin2312 pada mata pelajaran Kimia untuk jenjang Sekolah Menengah Atas

Mohon bantuannya untuk soal ini​
Mohon bantuannya untuk soal ini​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

Penjelasan:

Lain kali bilangin ke gurunya suruh siapin tabel nilai termodinamika (atau siapin sendiri terus upload gambar tabelnya)

a)

\Delta H^{\theta} (Mg_{(s)}) = 0 \;\dfrac{kJ}{\text{mol}}, \Delta H^{\theta} (Pb^{2+}_{(aq)}) = -1.7 \;\dfrac{kJ}{\text{mol}}, \Delta H^{\theta} (Mg^{2+}_{(aq)}) = -466.9\;\dfrac{kJ}{\text{mol}}, \\\\\Delta H^{\theta} (Pb_{(s)}) = 0\;\dfrac{kJ}{\text{mol}}\\

\Delta H^{\theta}_{total} = \sum( \Delta H^{\theta}_{produk} -\Delta H^{\theta}_{reaktan} ) \\\Delta H^{\theta}_{total} = \Delta H^{\theta} (Mg^{2+}_{(aq)}) + \Delta H^{\theta} (Pb_{(s)}) - (\Delta H^{\theta} (Mg_{(s)})+\Delta H^{\theta} (Pb^{2+}_{(aq)}))\\\Delta H^{\theta}_{total} = -466.9 + 1.7\\\Delta H^{\theta}_{total} = -465.2 \;\dfrac{kJ}{\text{mol}}\\

\Delta S^{\theta} (Mg_{(s)}) = 32.68 \;\dfrac{kJ}{\text{mol}}, \Delta S^{\theta} (Pb^{2+}_{(aq)}) = 10.5 \;\dfrac{kJ}{\text{mol}}, \Delta S^{\theta} (Mg^{2+}_{(aq)}) = -138.1\;\dfrac{kJ}{\text{mol}}, \\\\\Delta S^{\theta} (Pb_{(s)}) = 64.81\;\dfrac{kJ}{\text{mol}}\\

\Delta S^{\theta}_{total} = \sum( \Delta S^{\theta}_{produk} -\Delta S^{\theta}_{reaktan} ) \\ \\\Delta S^{\theta}_{total} = \Delta S^{\theta} (Mg^{2+}_{(aq)}) + \Delta S^{\theta} (Pb_{(s)}) - (\Delta S^{\theta} (Mg_{(s)})+\Delta S^{\theta} (Pb^{2+}_{(aq)}))\\\Delta S^{\theta}_{total} = -138.1 + 64.81 - (32.68+10.5) = -116.47 \;\dfrac{kJ}{\text{mol}}\\

\Delta G^{\theta}_{total} = \Delta H^{\theta}_{total} - (T_{celsius}+273)\cdot \Delta S^{\theta}_{total}\\\Delta G^{\theta}_{total} =-465.2 + 298\cdot 116.47\\ \boxed{\boxed{\Delta G^{\theta}_{total} = 34242.86\;\dfrac{kJ}{\text{mol}}}}

mencari nilai Kc :

\Delta G^{\theta}_{total} = -R(T_{celsius}+273)\cdot \ln(K_c) \to K_c = e^{^{\textstyle -\dfrac{\Delta G^{\theta}_{total} }{R\cdot (T_{celsius}+273)}}}\\\\\boxed{\boxed{K_c = e^{^{\textstyle -\dfrac{34242.86 }{0.008314\cdot 298}}} \approx 3.604\cdot 10^{-6081}}}

b)

\Delta H^{\theta} (Br_2_{(l)}) = 0 \;\dfrac{kJ}{\text{mol}}, \Delta H^{\theta} (I^{-}_{(aq)}) = -55.19\;\dfrac{kJ}{\text{mol}}, \Delta H^{\theta} (Br^{-}_{(aq)}) = -120.9\;\dfrac{kJ}{\text{mol}}, \\\\\Delta H^{\theta} (I_2_{(s)}) = 0\;\dfrac{kJ}{\text{mol}}\\

\Delta H^{\theta}_{total} = \sum( \Delta H^{\theta}_{produk} -\Delta H^{\theta}_{reaktan} ) \\ \Delta H^{\theta}_{total} = -(0-2\cdot 55.19- (-2\cdot 120.9 + 0))\\ \Delta H^{\theta}_{total} = -131.42 \;\dfrac{kJ}{\text{mol}}

\Delta S^{\theta} (Br_2_{(l)}) = 152.23 \;\dfrac{kJ}{\text{mol}}, \Delta S^{\theta} (I^{-}_{(aq)}) = 11.13\;\dfrac{kJ}{\text{mol}}, \Delta S^{\theta} (Br^{-}_{(aq)}) = 80.71\;\dfrac{kJ}{\text{mol}}, \\\\\Delta S^{\theta} (I_2_{(s)}) = 116.14\;\dfrac{kJ}{\text{mol}}\\

\Delta S^{\theta}_{total} = \sum( \Delta S^{\theta}_{produk} -\Delta S^{\theta}_{reaktan} ) \\\Delta S^{\theta}_{total} = 2\cdot 80.71 + 116.14 - (152.23 + 2\cdot 11.13) = 103.07 \;\dfrac{kJ}{\text{mol}}\\

\Delta G^{\theta}_{total} = \Delta H^{\theta}_{total} - (T_{celsius}+273)\cdot \Delta S^{\theta}_{total}\\\Delta G^{\theta}_{total} = -131.42 - 298\cdot 103.07 \\ \boxed{\boxed{\Delta G^{\theta}_{total} = -30846.28\;\dfrac{kJ}{\text{mol}}}}

mencari nilai Kc :

K_c = e^{^{\textstyle -\dfrac{\Delta G^{\theta}_{total} }{R\cdot (T_{celsius}+273)}}}\\\\\boxed{\boxed{K_c = e^{^{\textstyle \dfrac{30846.28 }{0.008314\cdot 298}}} \approx 1.136\cdot 10^{5407}}}

Semoga dengan pertanyaan yang sudah terjawab oleh Yorozuya1234 dapat membantu memudahkan mengerjakan soal, tugas dan PR sekolah kalian.

Apabila terdapat kesalahan dalam mengerjakan soal, silahkan koreksi jawaban dengan mengirimkan email ke yomemimo.com melalui halaman Contact

Last Update: Fri, 23 Sep 22
<< Previous Post
Next Post >>