Chemistry, 20.06.2020 12:57 stormserena

A 0.56 g sample of liquid C6H12 was combusted completely using excess oxygen inside a bomb (constant volume) calorimeter, with the products being carbon dioxide and liquid water. The calorimeter's heat capacity is 5.36 kJ °C-1. If the temperature inside the calorimeter increased from 25.0 °C to 29.2 °C, determine ΔrH for this reaction in kJ mol-1 (with respect to C6H12) at 298 K. Do not worry about how realistic the final answer is. You have 5 attempts at this question. TIP: To report an answer in scientific notation, enter it using the format "2.3E4", which means "2.3 x 104" (without the quotation marks)

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Answer from: lillianradecki

-3.4 × 10³ kJ/mol

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

$Qcal = c_{cal} \times \Delta T$

where,

Qcal : heat absorbed by the calorimeter

$c_{cal}$ : heat capacity of the calorimeter

$\Delta T$ : change in the temperature

$Qcal = \frac{5.36kJ}{\° C} \times (29.2\° C - 25.0 \° C ) = 22.5 kJ$

Step 2: Calculate the heat released by the combustion

According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the combustion is zero.

$Q_{cal} + Q_{comb} = 0\\Q_{comb} = -Q_{cal} = -22.5 kJ$

Step 3: Calculate the standard enthalpy of the combustion

22.5 kJ are released when 0.56 g of C₆H₁₂ (MW 84.16) is combusted. The standard enthalpy of the combustion (ΔH°c) is:

$\frac{-22.5kJ}{0.56g} \times \frac{84.16g}{mol} = -3.4 \times 10^{3} kJ/mol$

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

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

$\delta h=e_p-e_r$

$\delta h$ = enthaply

$e_p$ = energy of products

$e_r$ = energy of reactants

as $e_p< e_r$ , thus $\delta h$ comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

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