Answer:
K = 4.07x10⁻³
Explanation:
Based on the reaction:
NH₄I(s) ⇄ NH₃(g) + HI(g)
You can define K of equilibrium as the ratio of concentrations of reactants and products, thus:
K = [NH₃] [HI] / [NH₄I]
But, as NH₄I is a solid, is not taken into account in the equilibrium, that means K expression is:
K = [NH₃] [HI]
As the concentrations in equilibrium of the gases is:
[NH₃] = 4.34x10⁻²M
[HI] = 9.39x10⁻²M
Equilibrium constant, K, is:
K = 4.34x10⁻²M * 9.39x10⁻²M
<h3>K = 4.07x10⁻³</h3>
Ammonium hydroxide is basic
∆H° of the following reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
<h3>What is Bond Enthalpy? </h3>
The minimum amount of energy which is required to braak down or form the bonds in chemical reaction is known as bond enthalpy.
It can be calculated as:
∆Hrxn = sum of ∆H bond broken - sum. of ∆H of bond formed.
In order to Calculate ∆Hrxn for the given equation we have:
Bond energies in kJ/mol
- H—H = 436
- H—I = 295
- I—I = 151
Now, the given reaction is
H₂(g) + I₂(g) → 2HI(g)
Here, 1 mol of H₂ and 1 mole of I₂ breaks to form 2 moles of HI.
Therefore,
We know that,
∆Hrxn = B. E(H—H) + B. E(I—I) - 2B. E(H—I)
= 436 + 151 - 2× 295
= 436+ 151 - 590
∆Hrxn = -3kJ/mol.
Thus, from the above conclusion we can say that ∆Hrxn of the reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
learn more about Bond energy:
brainly.com/question/26964179
#SPJ4
Answer:
To calculate the number we need molar mass.