Kontes Terbuka Olimpiade Kimia November 2018

Diskusi bidang kimia

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Wed Nov 28, 2018 4:01 pm

Silahkan gunakan tempat ini untuk berdiskusi seputar soal-soal KTO Kimia :)

Soal dapat dilihat di: Link

Ralat soal untuk Metabolisme Glukosa, reaksi seharusnya:

$\ce {Glukosa + O2 + H2O -> Asam Glukonat + H2O2}$

Sehingga apabila anda menjawab asam glukonat dengan rumus $\ce{C6H10O6}$ akan kami benarkan.
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Thu Dec 06, 2018 1:44 am

mohon koreksinya kalo ada salah :)
Kasus Kopi Sianida
  1. KCN
  2. K = +1
    $\ce{ [^-:C#N:] }$
    $\ce{ \begin{align*}mf C &= 4 - \frac{1}{2}*6 - 2 = -1 \\ mf N &= 5 - \frac{1}{2}*6 - 2 = 0\end{align*} }$
  3. .
    $\ce{ \begin{align*}m_{KCN} &= \frac{M_r KCN}{M_r CN^-}*D_{50} \space CN^-*body mass \\ &= \frac{65.12 g mol^{-1}}{26.02 g mol^{-1}}*1.52 mg {kg}^{-1}*55.0 kg \\ \therefore m_{KCN} &= 209.2 mg\end{align*} }$
  4. .
    $\ce{ \begin{align*}[CN-] = [K+] = [KCN] &= \frac{n_{KCN}}{V_{kopi}} \\ &= \frac{0.209 g}{65.12 g mol^{-1}*0.150 L} \\ \therefore [CN-] &= 0.0214 mol L^{-1}\end{align*} }$
  5. $\ce{ \begin{align*}HCN ${(aq)}$ &<=> H+ ${(aq)}$ + CN- ${(aq)}$ &K_a = 10^{-11}\end{align*} }$
  6. .
    $\ce{ \begin{align*}K_a &= \frac{[H+][CN-]}{[HCN]} \\ [CN-] &= \frac{K_a[HCN]}{[H+]} \\ C_{0 \space (HCN)} &= [HCN] + [CN-] \\ &= [HCN](\frac{[H+] + K_a}{[H+]}) \\ [HCN] &= \frac{[H+]*C_{0 \space (HCN)}}{[H+] + K_a} \\ [CN-] &= \frac{K_a*C_{0 \space (HCN)}}{[H+] + K_a}\end{align*} }$
    $\ce{ \begin{align*}[K+] &= [OH-] + [CN-] \\ [K+] &= \frac{K_w}{[H+]} + \frac{K_a*C_{0 \space (HCN)}}{[H+] + K_a} \\ &= \frac{K_w*K_a + [H+](K_a*C_{0 \space (HCN)} + K_w)}{[H+]^2 + K_a*[H+]} \\ 0 &= [K+][H+]^2 + K_a([K+] - (C_{0 \space (HCN)} + K_w))[H+] - K_w*K_a \\ 0 &= 0.0214[H+]^2 + 10^{-11}(0.0214 - (0.0214 + 10^{-14}))[H+] - 10^{-25} \\ \therefore [H+] &= 2.16*10^{-12} M \\ \therefore {pH} &= 11.7\end{align*} }$
Proses Haber-Bosch
  1. $\ce{ \begin{align*}N2 ${(g)}$ + 3H2 ${(g)}$ &<=> 2NH3 ${(g)}$ &\Delta{H}^o_R = -91.6 {kJ} mol^{-1}\end{align*} }$
  2. Tekanan tinggi agar reaksi bergeser ke arah produk. Karena entalpi reaksi negatif (reaksi melepas kalor), sebaiknya dilangsungkan pada temperatur rendah.
  3. $\ce{ K_p = \frac{P_{NH_3}^2}{P_{N_2}*P_{H_2}^3} }$
  4. .
    $\ce{ \begin{align*} K_{400} &= K_{300}*e^{- \frac{\Delta{H}^o_R}{R}(\frac{1}{400} - \frac{1}{300})} \\ &= 4.34*10^{-3}*e^{\frac{91.6*10^3 \space J \space mol^{-1}}{8.314 \space J \space mol^{-1} \space K^{-1}}(\frac{1}{400} - \frac{1}{300}) K^{-1}} \\ &= 4.47*10^{-7}\end{align*} }$
  5. Karena terjadi pemutusan ikatan $\ce{ N#N }$ yang terdiri atas 1 ikatan $\ce{ \sigma }$ dan 2 ikatan $\ce{ \pi }$
  6. .
    $\ce{ \begin{align*} \frac{d[H{*}]}{dt} &= $k_1$[H2]\theta^2 - $k_4$[N{*}][H{*}]^3 = 0 \\ $k_4$[N{*}][H{*}]^3 &= $k_1$[H2]\theta^2 \\ \frac{d[N{*}]}{dt} &= $k_3$[N2{*}]\theta \space - $k_4$[N{*}][H{*}]^3 = 0 \\ $k_4$[N{*}][H{*}]^3 &= $k_3$[N2{*}]\theta \\ \frac{d[N2{*}]}{dt} &= $k_2$[N2]\theta \space - $k_3$[N2{*}]\theta = 0 \\ [N2{*}] &= \frac{$k_2$[N2]}{$k_3$} \\ \therefore \theta &= \frac{$k_3$[N2{*}]}{$k_1$[H2]} = \frac{$k_2$[N2]}{$k_1$[H2]} \\ \frac{d\theta}{dt} &= -$k_1$[H2]\theta^2 - $k_2$[N2]\theta \space - $k_3$[N2{*}]\theta \space + $k_4$[N{*}][H{*}]^3 + $k_5$[NH3{*}] = 0\end{align*} }$
    dengan menyubstitusikan persamaan-persamaan di atas, kita mendapatkan
    $\ce{ $k_5$[NH3{*}] = \frac{2$k_2$[N2]^2}{$k_1$[H2]} }$
    $\ce{ \begin{align*} \frac{d[NH3{*}]}{dt} &= $k_4$[N{*}][H{*}]^3 - $k_5$[NH3{*}] = 0 \\ r &= $k_4$[N{*}][H{*}]^3 = $k_5$[NH3{*}] \\ \therefore r &= \frac{2$k_2$[N2]^2}{$k_1$[H2]}\end{align*} }$
Kalorimetri
  1. .
    $\ce{ \begin{align*} Q_1 &= Q_2 \\ \Delta{T}_1(m*c + C) &= m*c*\Delta{T}_2 \\ C &= m*c(\frac{\Delta{T}_2}{\Delta{T}_1} - 1) \\ &= 50 {mL}*1 g {mL}^{-1}*4.2 J g^{-1} ^oC^{-1}(\frac{70 - 42 ^oC}{42 - 25 ^oC} - 1) \\ \therefore C &= 135.88 J ^oC^{-1}\end{align*} }$
  2. .
    $\ce{ \begin{align*}50 {mL} &-> 9.148 g \\ 10 {mL} &-> 1.830 g\end{align*} }$
    $\ce{ \begin{align*}\underset{$\frac{x}{36.5}$}{HCl} + NaOH &-> NaCl + \underset{$\frac{x}{36.5}$}{H2O} &\Delta{H}_1 = $\frac{-25.2x}{36.5} = - 0.690x$ {kJ} \\ \frac{1}{2}\underset{$\frac{1.83 - x}{98}$}{H2SO4} + NaOH &-> \frac{1}{2}Na2SO4 + \underset{$\frac{1.83 - x}{49}$}{H2O} &\Delta{H}_2 = $\frac{-53.3(1.83 - x)}{49} = 1.09x - 2$ {kJ} \\ HCl + \frac{1}{2}H2SO4 + 2NaOH &-> NaCl + \frac{1}{2}Na2SO4 + \underset{$\frac{12.5x + 66.8}{1788.5}$}{2H2O} &\Delta{H}_3 = $0.4x - 2$ {kJ}\end{align*} }$

    asumsi semua NaOH bereaksi
    $\ce{ \begin{align*} Q &= Q \\ -\Delta{H}_3 &= \Delta{T}_{larutan}(m*c + C) \\ 2 - $0.4x$ {kJ} &= 2.0^oC(50 {mL}*1 g {mL}^{-1}*4.2 J g^{-1} ^oC^{-1} + 135.88 J ^oC^{-1})*\frac{1 {kJ}}{1000 J} \\ \therefore $x$ &= 3.27 g\end{align*} }$
nah ini ane bingung gan, soalnya $\ce{ $x$ }$-nya lebih dari 1.83
mohon pencerahannya
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