Question

Consider the reaction \[ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \] Using standard thermodynamic data at \( 298 \mathrm{~K} \), calculate the entropy change for the surroundings when 2.44 moles of \( \mathrm{H}_{2}(\mathrm{~g}) \) react at standard conditions. \[ \Delta \mathrm{S}^{\circ} \text { surroundings }= \] \( \square \) \( \mid J / K \)

          Consider the reaction
\[
\mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})
\]

Using standard thermodynamic data at \( 298 \mathrm{~K} \), calculate the entropy change for the surroundings when 2.44 moles of \( \mathrm{H}_{2}(\mathrm{~g}) \) react at standard conditions.
\[
\Delta \mathrm{S}^{\circ} \text { surroundings }=
\]
\( \square \) \( \mid J / K \)

Consider the reaction

H2( g)+C2H4( g) ⟶C2H6( g)

Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 2.44 moles of H2( g) react at standard conditions.

ΔS^∘ surroundings =

□ | J / K

Added by Andrew C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Oluwapelumi Kolawole

Step 1

The entropy change for the surroundings during a chemical reaction at constant temperature and pressure can be calculated using the equation: \[ \Delta S^{\circ}_{\text{surroundings}} = -\frac{\Delta H^{\circ}_{\text{reaction}}}{T} \] where \(\Delta Show more…

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