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The following reaction describes the breakdown of glucose in your body:

$$C_{6}H_{12}O_{6}(s) + 6O_{2}(g) → 6CO_{2}(g) + 6H_{2}O(l)$$

$\Delta H_{f}^{\circ}\space kJ/mol$ $S^{\circ}J/mol K$
$C_{6}H_{12}O_{6}$ $-1274.4$ $212.1$
$O_{2}$ $0$ $205.0$
$H_{2}O$ $-285.8$ $69.91$
$CO_{2}$ $-393.5$ $213.6$

$\ $
What is the equilibrium constant for this reaction at body temperature ($37^{\circ}C$)?

Assume that enthalpies of formation and absolute entropies are temperature independent.

A

$0.33$

B

$3.1$

C

$5.6 \times 10^{13}$

D

$1.7 \times 10^{-14}$

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