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VashaNatasha [74]
3 years ago
13

Without a catalyst, a reaction may require very high temperatures in order to occur. How does a catalyst speed a reaction at low

temperatures? It raises the amount of energy required to break chemical bonds. It lowers the amount of energy required to break chemical bonds. It increases the concentration of the reactant particles. It decreases the concentration of the reactant particles.
Chemistry
2 answers:
Wittaler [7]3 years ago
7 0

It lowers the amount of energy required to break chemical bonds.

Marrrta [24]3 years ago
4 0

The correct answer is "It lowers the amount of energy required to break chemical bonds."

Catalyst work by decreasing the activation energy of a chemical reaction. Activation energy is the minimum energy required to start a chemical reaction. For a chemical reaction to start the bonds in the reactant molecules are broken and new bonds are formed in the product molecules. Thus catalyst facilitate to break chemical bonds in reactants  by lowering the activation energy.

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The gas cyclobutane, C4H8(g), can be used in welding. When cyclobutane is burned in oxygen, the reaction is: C4H8(g) + 6 O2(g)4
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Answer:

a

\Delta H^o _{rxn} = -2568.9 \  kJ

b

H  = 350 JK^{-1}

c

T_{max}  = 32.4 ^o C

Explanation:

From the question we are told that

 The reaction of cyclobutane and oxygen is

         C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}

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

      \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)}) ] ]

=>  \Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]

=>  \Delta H^o _{rxn} = -2568.9 \  kJ

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

     H  = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]

=>  H  = [ 4 * 37.1 + 6* 33.6 ]

=>   H  = 350 JK^{-1}

From the question the initial temperature of reactant is  T_i  =  25^oC

Generally the enthalpy change(\Delta H^o _{rxn}) of the reaction is mathematically represented as

 |\Delta H^o _{rxn} |=  H  * (T_{max} -T_i)

  2568.9 =   350  * (T_{max} -25)  

=> \frac{2568.9 }{350}  =  T_{max} - 25

=> T_{max}  = 32.4 ^o C

   

4 0
3 years ago
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