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  • Consider the reaction: <img alt="\left(\mathrm{CH}_3\right)_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{CH}_3 \mathrm{OH}(\mathrm{l})" src="https://

Consider the reaction:

\left(\mathrm{CH}_3\right)_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{CH}_3 \mathrm{OH}(\mathrm{l})

Given the standard enthalpy change (H) for the reaction at 298 K is -64.5 kJ/mol, calculate the enthalpy change when 2 moles of \mathrm{(CH_{3})_{2}O} and 1 mole of \mathrm{H_{2}O} react to form 2 moles of  \mathrm{CH_{3}OH}.

Option: 1

\mathrm{\Delta H=-64.5 \mathrm{~kJ}}


Option: 2

\mathrm{\Delta H=-129.0 \mathrm{~kJ}}


Option: 3

\mathrm{\Delta H=-32.3 \mathrm{~kJ}}


Option: 4

\mathrm{\Delta H=-258.0 \mathrm{~kJ}}


Answers (1)

best_answer

Step 1: Write the Reaction and Given Information The given reaction is:

\left(\mathrm{CH}_3\right)_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{CH}_3 \mathrm{OH}(\mathrm{l})

Given standard enthalpy change (?H?) of the reaction at 298 K = -64.5 kJ/mol.

Reacting moles: 2 moles of \left(\mathrm{CH}_3\right)_2 \mathrm{O} and 1 mole of \mathrm{H}_2 \mathrm{O} .

Step 2: Calculate Enthalpy Change\mathrm{(\Delta I I)}  for Given Reactant Moles Calculate the enthalpy change (?H) for the given reactant moles us- ing the given standard enthalpy change\mathrm{(\Delta H\degree)} and the reaction stoichiometry:

 \mathrm{\begin{gathered} \Delta H=\Delta H^{\circ} \times(\text { molcs of rcaction }) \\ \Delta H=-64.5 \mathrm{~kJ} / \mathrm{mol} \times 1=-64.5 \mathrm{~kJ} \end{gathered}}

Step 3: Compare ?H with Options Comparing the calculated ?H = −64.5 kJ with the provided options, we see that the correct option is:
Correct Answer: A) ?H = −64.5 kJ

Posted by

Divya Prakash Singh

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