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Consider the following reaction, for which ∆H° = +87.9 kJ/mol and ∆S° = +170.2 J/mol·K. PCl5(g) → PCl3(g) + Cl2(g) Assume ∆H° and ∆S° do not vary appreciably with change in temperature. At what temperature will the reaction be at equilibrium when the partial pressures of each component of the reaction are under standard conditions (1.00 atm each).?

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pstnonsonjoku

Answer:

516.45K or 243.45°C

Explanation:

From

∆S= ∆H/T

∆S= change in entropy

∆H= change in enthalpy

T= temperature in Kelvin

T= ∆H/∆S

T= 87.9 ×10^3Jmol-1 / 170.2Jmol-1K-1

T= 516.45K or 243.45°C

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