Answer:
a
[tex]\Delta H^o _{rxn} = -2568.9 \ kJ [/tex]
b
[tex]H = 350 JK^{-1}[/tex]
c
[tex]T_{max} = 32.4 ^o C[/tex]
Explanation:
From the question we are told that
The reaction of cyclobutane and oxygen is
[tex]C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}[/tex]
ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ
Generally ΔH° for this reaction is mathematically represented as
[tex]\Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ][/tex]
=> [tex]\Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0][/tex]
=> [tex]\Delta H^o _{rxn} = -2568.9 \ kJ [/tex]
Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is mathematically represented as
[tex]H = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}][/tex]
=> [tex]H = [ 4 * 37.1 + 6* 33.6 ][/tex]
=> [tex]H = 350 JK^{-1}[/tex]
From the question the initial temperature of reactant is [tex]T_i = 25^oC[/tex]
Generally the enthalpy change([tex]\Delta H^o _{rxn}[/tex]) of the reaction is mathematically represented as
[tex]|\Delta H^o _{rxn} |= H * (T_{max} -T_i)[/tex]
[tex] 2568.9 = 350 * (T_{max} -25)[/tex]
=> [tex]\frac{2568.9 }{350} = T_{max} - 25[/tex]
=> [tex]T_{max} = 32.4 ^o C[/tex]
