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pickupchik [31]
3 years ago
7

HELP ASAP! Chemical reactions involving liquids only will be influenced by ALL of the following EXCEPT: a) temperature of the sy

stem b) number of reactant particles c) size of reactant particles d) pressure of the system
Chemistry
1 answer:
Rom4ik [11]3 years ago
8 0
<span>Chemical reactions involving only liquids will be influenced by ALL of the following EXCEPT: </span><span><span>c) size of reactant particles d) pressure of the system.

</span>Reason:
1) As per Arrhenius theory, rate of chemical reactions is influence by temperature. Mathematically it is expressed as: k = A exp(-Ea/RT)
</span>where, k = rate constant, A = collision frequency, Ea = activation energy, T = temp. From above relation, it can be seen that, chemical reactions involving only liquids will be influenced temperature.

2) Also, if know that, rates of chemical reactions is mathematically expressed as: Rate = [A]^{x} B^{y}....,
where A, B, .. are the reactants.
From above relation, it can be seen that, chemical reactions involving only liquids will be influenced concentration i.e. number of particle. 

3) However, since all the reactant and catalyst used (if any) is in liquid state, particle size of same  will not influence the reaction.

4) Also, since there are no gas-phase reactant, pressure will not affect the reaction. 
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Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

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