Answer to Question #198891 in General Chemistry for sarah

Question

The Neches Butane Products Co. makes butadiene, C,Hg. by separating butenes. C,Hg. from refinery gases piped to them, and then further dehydrogenating the butenes to butadiene. In order to increase capacity they would like to be able to dehy- drogenate butane, C,H10, in one step without the intermediate separation of the butene. The CH formed would be separated, and any unreacted C,H10 or C,H would be recycled to the reactor. One of the large engineering companies claims to have such a process. You have just been employed by Neches B.P. Co.-fresh from school-and are requested to make some calculations on the proposed process: CH,p(1) C,H(g) + 2H,(g) By heat exchange, the butane fed into the reactor is brought to the reaction tempera- ture of 1000 K. It is desired to maintain an isothermal reactor. How much heat must be added or removed per g mole of butadiene formed? Assume the pressure is 100 kPa. What are some reasons why the heat load may be different from the value you calculate?

Answer 1

$C_4H_{10}(l)\longrightarrow C_4H_6(g)+2H_2(g)$

$\Delta H_R^o$ = Heat of reaction at 298.15 K

$=1\times(-165.5\space kJ/g)-1\times(-147\space kJ/g)$

$=-18.5\space kJ/g$

The standard heat of reaction at any temperature T is given by,

$\Delta H_{RT}^o=\Delta H_o+aT+bT^2+cT^3+dT^4$

$\Delta H_{RT}^o=\Delta H_o+11.92T+1.35\times10^{-1}T^2-0.00004767T^3+7.368\times10^{-9}T^4$

$\Delta H_T$ at 298 K = $-18.5\space kJ/g$

$-18.5=\Delta H_o+11.92\times298+1.35\times10^{-1}(298)^2-0.00004767\times(298)^3+7.368\times10^{-9}\times (298)^4$

$\Delta H_o=14337.28\space kJ/g$

$\Delta H^o_{RT}$ at T = 1000 K

$\Delta H_{RT}^o=\Delta H_o+11.92T+1.35\times10^{-1}T^2-0.00004767T^3+7.368\times10^{-9}T^4$

$\Delta H_{RT}^o=5.81\times10^{10}+11.92\times1000+1.35\times10^{-1}(1000)^2-0.00004767(1000)^3+7.368\times10^{-9}(1000)^4$

$\Delta H_{RT}^o=120,955\space kJ/g$

Amount of heat needed to be added is equal to 120,955 kJ/g