© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Reactions
We are given two combustion reactions:
1) For methane ($CH_4$):
$$CH_4 + 2\,O_2 \to CO_2 + 2\,H_2O,\quad \Delta H = x$$
2) For methanol ($CH_3OH$):
$$CH_3OH + \tfrac{3}{2}\,O_2 \to CO_2 + 2\,H_2O,\quad \Delta H = y$$
Additionally, the reaction for the partial oxidation of methane to methanol is:
$$CH_4 + \tfrac{1}{2}\,O_2 \to CH_3OH,\quad \Delta H_r < 0 \quad (\text{given})$$
Step 2: Relate Enthalpy Changes
We can obtain the partial oxidation reaction by subtracting the methanol combustion reaction from the methane combustion reaction:
$$\bigl[\,CH_4 + 2\,O_2 \to CO_2 + 2\,H_2O\bigr] - \bigl[\,CH_3OH + \tfrac{3}{2}\,O_2 \to CO_2 + 2\,H_2O\bigr]$$
This subtraction simplifies to:
$$CH_4 + \tfrac{1}{2}\,O_2 \to CH_3OH$$
Hence, the enthalpy of the partial oxidation reaction is:
$$\Delta H_r = x - y$$
Step 3: Analyze the Sign of the Enthalpy Change
We are given that this partial oxidation reaction's enthalpy change ($\Delta H_r$) is negative:
$$\Delta H_r = x - y < 0$$
Therefore:
$$x - y < 0 \quad \Rightarrow \quad x < y$$
Step 4: Conclude the Correct Relationship
Since $x < y$, the correct answer to the given question is:
x < y
