Calculate the value of enthalpy change at 300 K of the reaction if equilibrium constant as a function of temperature is given as K_{eq} = 40 e^{\frac{- \textrm{ } 40}{T}} \textrm{ } \cdot \textrm{ } e^{\frac{300}{T^{2}}} .

Engineering

Chemistry

Arrhenius Equation

Question

Calculate the value of enthalpy change at 300 K of the reaction if equilibrium constant as a function of temperature is given as

$K_{eq} = 40 e^{\frac{- 40}{T}} \cdot e^{\frac{300}{T^{2}}}$ .

38 R

300 R

40 R

80 R

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$K = 30 e^{\frac{- 40}{T}} \cdot e^{\frac{300}{T^{2}}}$

$lnK = \ln 30 - \frac{40}{T} + \frac{300}{T^{2}}$
$\frac{lnK}{dT} = \frac{E_{0}}{{RT}^{2}}$
$\frac{E_{0}}{R} = 40 - \frac{300 \times 2}{T}$

$\frac{E °}{R} = 40 - \frac{600}{300}$

$\frac{E °}{R} = 38$

E° = 38 × R