Calcule a entalpia padrão de reacão, na temperatura de 800 K , para a reação:
\hat{\mathrm{C}}_{131}+\mathrm{H}_{2} \mathrm{O}_{19} \rightarrow \mathrm{CO}_{(8)}-\mathrm{H}_{2(9)} \quad \mathrm{AH} \mathrm{C}_{24 \mathrm{G}_{1} \mathrm{O}}=+131,24 \mathrm{~K}
Sendo conhecidas as seguintes capacidades térmicas,
\begin{aligned} & \mathrm{C}_{p}{ }^{0}\left[\mathrm{CO}_{(\mathrm{g})}\right]=26,86+6,97 \times 10^{-3} \mathrm{~T}-8,20 \times 10^{-7} \mathrm{~T}^{2} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\\ & \mathrm{C}_{p}{ }^{0}\left[\mathrm{H}_{2(\mathrm{~g})}\right]=29,07+8,36 \times 10^{-4} \mathrm{~T}+2,01 \times 10^{-6} \mathrm{~T}^{2} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\\ & \mathrm{C}_{p}{ }^{0}\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]=27,87+9,61 \times 10^{-3} \mathrm{~T}+1,18 \times 10^{-6} \mathrm{~T}^{2} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\\ & \mathrm{C}_{p}{ }^{0}\left[\mathrm{C}_{(\mathrm{s})}\right]=-5,27+5,86 \times 10^{-2} \mathrm{~T}-4,32 \times 10^{-5} \mathrm{~T}^{2}+1,15 \times 10^{-8} \mathrm{~T}^{3} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}

Passo 1

O enunciado pede a entalpia padrão de reação a 800\\ \mathrm{K} para
\mathrm{C_{(s)}} + \mathrm{H_{2}O_{(g)}} \longrightarrow \mathrm{CO_{(g)}} + \mathrm{H_{2(g)}}
sabendo que a entalpia padrão de reação a 298\\ \mathrm{K} é \Delta H_{298}^\circ=+131{,}24\\ \mathrm{kJ/mol} e dadas as capacidades térmicas molares C_p^0(T) de cada espécie.
Conteúdo: Lei de Kirchhoff e integração de polinômios de C_p(T).

Passo 2

Pela lei de Kirchhoff:
\Delta H^\circ(800) \;=\;\Delta H_{298}^\circ \;+\;\int_{298}^{800}\Bigl[\sum C_p^0(\text{produtos})-\sum C_p^0(\text{reagentes})\Bigr]\,dT.

Passo 3

Calculemos o polinômio
\Delta C_p^0(T) = C_p^0(\mathrm{CO}) + C_p^0(\mathrm{H_2}) - C_p^0(\mathrm{H_2O}) - C_p^0(\mathrm{C}) = a_0 + a_1 T + a_2 T^2 + a_3 T^3
com coeficientes (em \mathrm{J\,K^{-1}\,mol^{-1}}):
\begin{aligned} a_0 &=26{,}86+29{,}07-27{,}87-(-5{,}27)=33{,}33,\\\ a_1 &=6{,}97\times10^{-3}+8{,}36\times10^{-4}-9{,}61\times10^{-3}-5{,}86\times10^{-2} =-6{,}0404\times10^{-2},\\\ a_2 &=-8{,}20\times10^{-7}+2{,}01\times10^{-6}-1{,}18\times10^{-6}+4{,}32\times10^{-5} =4{,}321\times10^{-5},\\\ a_3 &=0+0+0-1{,}15\times10^{-8}=-1{,}15\times10^{-8}. \end{aligned}

Passo 4

Integramos termo a termo de 298 a 800\\ \mathrm{K}:
\int_{298}^{800}\Delta C_p^0\,dT = a_0\,(800-298) + \frac{a_1}{2}\bigl(800^2-298^2\bigr) + \frac{a_2}{3}\bigl(800^3-298^3\bigr) + \frac{a_3}{4}\bigl(800^4-298^4\bigr).

Passo 5

Calculando cada parcela (em \mathrm{J/mol}):
\begin{aligned} a_0\,(800-298)&=33{,}33\times502=1,67\times10^4,\\\ \frac{a_1}{2}(800^2-298^2)&=-0{,}030202\times551196\approx-1,66\times10^4,\\\ \frac{a_2}{3}(800^3-298^3)&=1{,}4403\times10^{-5}\times4{,}855\times10^8\approx+7{,}01\times10^3,\\\ \frac{a_3}{4}(800^4-298^4)&=-2{,}875\times10^{-9}\times4{,}017\times10^{11}\approx-1{,}15\times10^3. \end{aligned}
Somando:
\int_{298}^{800}\Delta C_p^0\,dT\approx5\,937\\ \mathrm{J/mol}.

Passo 6

Convertemos para \mathrm{kJ/mol} e aplicamos Kirchhoff:
\Delta H^\circ(800) =131{,}24\\ \mathrm{kJ/mol} + \frac{5{,}937\\ \mathrm{J/mol}}{1000} \approx131{,}24 +5{,}94 =137{,}18\\ \mathrm{kJ/mol}.

Resposta

\displaystyle \Delta H^\circ(800\\ \mathrm{K})\approx +137{,}2\\ \mathrm{kJ/mol}.

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