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stepladder [879]
3 years ago
9

A 35.6 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ros

e from 35.0 to 76.0°C and the heat capacity of the calorimeter is
23.3 kJ/°C, what is the value of DH°rxn? The molar mass of ethanol is 46.07 g/mol.

C2H5OH(l) + O2(g) → CO2(g) + H2O(g) ΔH°rxn = ? (Points : 1)
-1.24 × 103 kJ/mol
+1.24 × 103 kJ/mol
-8.09 × 103 kJ/mol
-9.55 × 103 kJ/mol
+9.55 × 103 kJ/mol
Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
7 0
<h3>Answer:</h3>

1.24 × 10³ kJ/mol

<h3>Explanation:</h3>

From the question we are given;

Heat capacity of the calorimeter =23.3 kJ/°C

Temperature change, ΔT = 76°C - 35°C

                                          =  41 °C

Mass of ethanol = 35.6 g

Molar mass of ethanol = 46.07 g/mol

We are required to determine the molar enthalpy

We can use the following steps:

<h3> Step 1 : Calculate the heat change of the reaction</h3>

Heat change will be equivalent to heat gained by the calorimeter.

Therefore;

Heat = heat capacity × change in temperature

Q = CΔT

   =  23.33 kJ/°C × 41°C

   = 955.3 kJ

<h3>Step 2 : Calculate the moles of ethanol burned </h3>

Moles = mass ÷ Molar mass

Therefore;

Moles of ethanol = 35.6 g ÷ 46.07 g/mol

                            = 0.773 moles

<h3>Step 3: Calculate the molar enthalpy of the reaction </h3>

Heat change for 0.773 moles of ethanol is 955.3 kJ

0.773 moles = 955.3 kJ

1 mole will have ,

    = 955.3 kJ ÷ 0.773 moles

    = 1235.83 kJ/mol

    = 1.24 × 10³ kJ/mol

But since the reaction is exothermic (release of heat) then the enthalpy change will have a negative sign.

Thus;

ΔH = -1.24 × 10³ kJ/mol

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<u>Answer:</u>

(A)

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