Question 15.2.3: Manipulate equilibrium constant expressions. The equilibrium......
Manipulate equilibrium constant expressions.
The equilibrium constant for the reaction of hydrogen with oxygen to form water is 5.7 × 10^{40} at 25 ºC.
H_{2}(g) + ½ O_{2}(g) \rightleftarrows H_{2}O(g) K = 5.7 × 10^{40} at 25 ºC
Calculate the equilibrium constant for the following reactions.
a. 2 H_{2}(g) + O_{2}(g) \rightleftarrows 2 H_{2}O(g)
b. H_{2}O(g) \rightleftarrows H_{2}(g) + ½ O_{2}(g)
You are asked to calculate an equilibrium constant for a reaction.
You are given a chemical equation, with an equilibrium constant, that is related to the equation for the given reaction.
a. Multiplying the original equation by a factor of 2 results in the new equation, so the new equilibrium constant is equal to the original constant squared (raised to the power of 2).
K_{original} = \frac{\left[H_{2}O\right] }{\left[H_{2}\right]\left[O_{2}\right] ^{1/2} } K_{new} = \frac{\left[H_{2}O\right]^{2} }{\left[H_{2}\right]^{2}\left[O_{2}\right] } = K_{original}^{2}
K_{new} = K_{original}^{2} = (5.7 × 10^{40})^2 = 3.2 × 10^{81}
b. Reversing the original equation results in the new equation, so the new equilibrium constant is equal to the inverse of the original constant.
K_{original} = \frac{\left[H_{2}O\right] }{\left[H_{2}\right]\left[O_{2}\right] ^{1/2} } K_{new} = \frac{\left[H_{2}\right]\left[O_{2}\right]^{{1}/{2}}}{\left[H_{2}O\right] }=\frac{1}{K_{original}}
K_{new} = \frac{1}{K_{original}} = \frac{1}{5.7 \times 10^{40}}=1.8 × 10^{-41}