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3241004551 [841]
2 years ago
9

For a particular reaction, the change in enthalpy is â9kJmole and the activation energy is 13kJmole. The enthalpy change (ÎH) an

d activation energy (Ea) for the catalyzed reaction could be which of the following? a. â9kJmole,5kJmole. b. â9kJmole,19kJmole.c. â14kJmole,5kJmole.d. â14kJmole,â13kJmole.
Chemistry
1 answer:
Svetllana [295]2 years ago
4 0

Answer:

The answer is Option a, that is "−9kJmole,5kJmole".

Explanation:

Please find the complete question in the attached file.

In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.

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Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
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<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

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