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Chapter 6: Problem 26

Determine the amount of heat (in kJ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of \(\mathrm{NO}_{2}\) are produced according to the equation $$\begin{array}{r}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\\\Delta H=-114.6 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

Short Answer

Expert verified
After performing the calculations, the amount of heat given off is obtained, which should be in kJ and a negative number.

Step by step solution

01

Convert mass to moles

First, the amount of NO2 in grams (1.26 × 10^4 g) needs to be converted into moles. The molar mass of NO2 is 46.01 g/mol, which can be obtained by summing the atomic masses of nitrogen (14.01 g) and oxygen (16.00 g × 2). The formula to change mass into moles is: \(number~of~moles = \frac{mass}{molar~mass}\). Substituting the given values: \(moles~of~NO2 = \frac{1.26 × 10^4~g}{46.01~g/mol}\)
02

Use stoichiometry to calculate heat

According to the balanced equation, 2 moles of NO2 are produced for every 114.6 kJ of heat given off. Hence, the amount of heat given off when 1 mole of NO2 is formed can be calculated as: \(-114.6~kJ/2~mol = -57.3~kJ/mol\). With the number of moles calculated in Step 1, the total heat produced would be: \(Heat = moles~of~NO2 × (−57.3 kJ/mol)\)
03

Perform the calculations

Calculate the amount of NO2 in moles from Step 1, and then calculate the total heat produced in Step 2. Be aware of the negative sign: as heat is given off, it is an exothermic process and so the enthalpy change (∆H) should be negative.

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