Question

what is the change in enthalpy in kilojoules when 3.24 g of ch₃oh gas is completely reacted according to the following reaction? 2 ch₃oh(g) → 2 ch₄(g) + o₂(g) δh = 252.8 kj 3.24 g ch₃oh × ( )

Explanation:

Step1: Calculate the molar mass of \(CH_3OH\)

The molar mass of \(C = 12.01\space g/mol\), \(H=1.008\space g/mol\), \(O = 16.00\space g/mol\). For \(CH_3OH\), \(M=(12.01 + 4\times1.008+16.00)\space g/mol=32.05\space g/mol\)

Step2: Find the number of moles of \(CH_3OH\)

Using the formula \(n=\frac{m}{M}\), where \(m = 3.24\space g\) and \(M = 32.05\space g/mol\). So \(n=\frac{3.24\space g}{32.05\space g/mol}\)

Step3: Use the enthalpy change from the reaction

From the reaction \(2CH_3OH(g)\to2CH_4(g)+O_2(g)\space \Delta H = 252.8\space kJ\), for \(2\space mol\) of \(CH_3OH\) the enthalpy change is \(252.8\space kJ\). Let \(x\) be the enthalpy change for \(n\space mol\) of \(CH_3OH\). Using proportion \(\frac{x}{n}=\frac{252.8\space kJ}{2\space mol}\)

Combining the steps: \(\Delta H=3.24\space g\space CH_3OH\times\frac{1\space mol\space CH_3OH}{32.05\space g\space CH_3OH}\times\frac{252.8\space kJ}{2\space mol\space CH_3OH}\)

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Answer:

\(12.8\space kJ\)