If the equilibrium constants for the two reactions
2 HCl(g)
⇔
H2(g) + Cl2(g)
and
I2(g) + Cl2(g)
⇔
2 ICl(g)
are denoted K1 and K2 respectively, then the equilibrium constant for the reaction
2 HCl(g) + I2(g)
⇔
2 ICl(g) + H2(g) equals

Question

If the equilibrium constants for the two reactions
2 HCl(g) 
⇔
H2(g) + Cl2(g)
and
I2(g) + Cl2(g) 
⇔
 2 ICl(g)
are denoted K1 and K2 respectively, then the equilibrium constant for the reaction
2 HCl(g) + I2(g) 
⇔
 2 ICl(g) + H2(g) equals

BlackTom AI · Accepted Answer

To tackle the combination of reactions, I’ll inspect how the overall equation emerges from the given ones and how the equilibrium constants combine.
Option 1: (K1/K2)2. This would suggest that the two steps contribute inversely with respect to K2 and then squared, which doesn’t align with the straightforward combination of forward reactions here. Since both original processes contribute multiplicatively rather than dividing one by the other, squaring or taking a ratio isn’t warranted.
Option 2: (K1K2)2. This would imply that the overall equilibrium constant is the square of the product of K1 and K2. The squaring arises if we doubled the number of times each reaction occurs, but in the net equation we only add the two reactions once each (with proper cancellation of species), so no extra factor of 2 appears in the exponent. Therefore, this is too large by a factor of 2 in the exponent.
Option 3: K1K2/2. This would indicate a halving of the product K1K2, which would only occur if there were an extra dividing step or a partial cancellation that reduces the effective contribution, but the systematic combination of the two forward reactions as written leads to a direct multiplication without such a division.
Option 4: K1+K2. Equating the overall constant to a sum of constants is not how equilibrium constants combine. Equilibrium constants multiply when you add reactions (after canceling identical species), not add. A sum would correspond to some unrelated thermodynamic aggregation, not the stoichiometric combination at hand.
Option 5: K1K2. This matches the proper rule: when you add two reactions and cancel identical species, the overall equilibrium constant is the product of the individual constants. Here, adding the forward form of 2 HCl ⇌ H2 + Cl2 and the forward form of I2 + Cl2 ⇌ 2 ICl, and canceling a common Cl2, yields the net equation 2 HCl + I2 ⇌ 2 ICl + H2, whose equilibrium constant is K1K2.

If the equilibrium constants for the two reactions 2 HCl(g) ⇔ H2(g) + Cl2(g) and I2(g) + Cl2(g) ⇔ 2 ICl(g) are denoted K1 and K2 respectively, then the equilibrium constant for the reaction 2 HCl(g) + I2(g) ⇔ 2 ICl(g) + H2(g) equals 单项选择题

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