Question

determine the amount of heat, in kj, given off when 329 g of no₂ is produced according to the following equation:
2 no(g) + o₂(g) → 2 no₂(g)
δh = -114.6 kj/mol
report your answer to the nearest whole number, and be sure to include the correct sign.

Explanation:

Step1: Calculate the molar mass of \(NO_2\)

The molar mass of \(N = 14.01\space g/mol\) and \(O=16.00\space g/mol\). For \(NO_2\), \(M=(14.01 + 2\times16.00)\space g/mol=46.01\space g/mol\)

Step2: Calculate the number of moles of \(NO_2\)

Using the formula \(n=\frac{m}{M}\), where \(m = 329\space g\) and \(M = 46.01\space g/mol\). So \(n=\frac{329\space g}{46.01\space g/mol}\approx7.15\space mol\)

Step3: Calculate the heat released

From the balanced equation \(2NO(g)+O_2(g)\to2NO_2(g)\), \(\Delta H=- 114.6\space kJ/mol\) (per \(2\space mol\) of \(NO_2\)).
For \(n = 7.15\space mol\) of \(NO_2\), the heat \(q\) is calculated as follows:
Since for \(2\space mol\) of \(NO_2\), \(\Delta H=-114.6\space kJ\), for \(1\space mol\) of \(NO_2\), \(\Delta H=\frac{- 114.6}{2}\space kJ/mol=-57.3\space kJ/mol\)
\(q=n\times\Delta H\), substituting \(n = 7.15\space mol\) and \(\Delta H=-57.3\space kJ/mol\)
\(q=7.15\times(-57.3)\space kJ\approx - 410\space kJ\)

Answer:

\(-410\space kJ\)