Chemistry, 30.03.2020 20:24 jonmorton159

Consider the following reaction at 298 K. C ( graphite ) + 2 H 2 ( g ) ⟶ CH 4 ( g ) Δ H ∘ = − 74.6 kJ and Δ S ∘ = − 80.8 J / K Calculate the following quantities. Δ S sys = J/K Δ S surr = J/K Δ S univ = J/K Is this reaction spontaneous? yes no

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Answer from: longoriafaithe09

Answer :

Entropy of system = -80.8 J/K

Entropy of surrounding = 253.7 J/K

Entropy of universe = 172.9 J/K

The reaction is spontaneous.

The process is a spontaneous process.

Explanation :

The given chemical reaction is:

$C(graphite)+2H_2(g)\rightarrow CH_4(g)$

Entropy of reaction = $\Delta S^o$ = Entropy of system = -80.8 J/K

Now we have to calculate the entropy of surrounding.

Entropy of surrounding = $\frac{-\Delta H^o}{T}$

Entropy of surrounding = $-\frac{-75.6kJ}{298K}$

Entropy of surrounding = $\frac{75.6\times 1000J}{298K}$

Entropy of surrounding = 253.7 J/K

As, we know that:

Entropy of universe = Entropy of system + Entropy of surrounding

Entropy of universe = -80.8 J/K + (253.7 J/K)

Entropy of universe = 172.9 J/K

Now we have to calculate the Gibbs free energy.

As we know that,

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

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -74600 J

$\Delta S^o$ = standard entropy = -80.8 J/K

T = temperature of reaction = 298 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(-74600J)-(298K\times -80.8J/K)$

$\Delta G^o=-50521.6J=-50.5kJ$

A reaction to be spontaneous when $\Delta G$ A reaction to be non-spontaneous when $\Delta G0$

For the reaction to be spontaneous, the Gibbs free energy of the reaction $\Delta G$ is negative or we can say that the value of

As the value of $\Delta G$ is less than zero that means the reaction is spontaneous.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

the product of the pressure and volume is also constant

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Consider the following reaction at 298 K. C ( graphite ) + 2 H 2 ( g ) ⟶ CH 4 ( g ) Δ H ∘ = − 74.6 k...

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