Chemistry, 17.12.2019 21:31 colestout2993

Consider the reaction: m + 2hcl → mcl2 + h2

when 0.25 mol of the metal, m, reacted with an aqueous hcl solution (the hcl is in excess), the temperature of the solution rose because the reaction produced 7025 j of heat. what is ∆h in kj per mol of m for this reaction? (hint: is this reaction exothermic or endothermic? )

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Answer from: stefaniethibodeaux

The enthalpy change of the reaction is -28.1 kJ/mol

Explanation:

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

= amount of heat released = -7025 J

n = number of moles of metal M = 0.25 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-7025J}{0.25mol}=-28100J/mol=-28.1kJ/mol$

Conversion factor used: 1 kJ = 1000 J

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and is written on the reactant side and is considered as an endothermic reaction.When heat is released, the sign of heat is taken to be negative and is written on the product side and is considered as an exothermic reaction.

So, the chemical reaction becomes:

$M+2HCl\rightarrow MCl_2+H_2+28.1kJ$

Hence, the enthalpy change of the reaction is -28.1 kJ/mol

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

pie chart because it is in percents ( % )

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