Chemistry, 10.12.2019 04:31 sbudlove2838

Gfind the ∆hrxn for the reaction: 3c(s)+4h2(g) →c3h8(g) using these reactions with known ∆h’s: c3h8(g) + 5o2(g) → 3co2(g) + 4h2o(g) ∆h = −2043 kj c(s) + o2(g) → co2(g) ∆h = −393.5 kj 2h2(g) + o2(g) → 2h2o(g) ∆h = −483.6 kj

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Answer from: dvh

The $\Delta H^o_{rxn}$ for the reaction is -104.7 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the formation of propane follows:

$3C(s)+4H_2(g)\rightarrow C_3H_8(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$ $\Delta H_1=-2043kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$ ( × 3)

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$ $\Delta H_3=-483.6kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[3\times \Delta H_2]+[2\times \Delta H_3]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-(-2043))+(3\times (-393.5))+(2\times (-483.6))]=-104.7kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -104.7 kJ.

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

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Gfind the ∆hrxn for the reaction: 3c(s)+4h2(g) →c3h8(g) using these reactions with known ∆h’s: c3h...

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