How Do You Solve an Equilibrium Problem with Initial Reactants and No Products?

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To solve an equilibrium problem with initial reactants and no products, start by defining the initial concentrations of the reactants and products. For the reaction 2HI(g) ↔ H2(g) + I2(g) at 527˚C with K = 0.0160, the initial concentrations are [H2] = [I2] = 0.0160 M and [HI] = 0 M. At equilibrium, let the change in concentration of HI be x, leading to [H2] = [I2] = 0.0160 - (x/2). The equilibrium constant expression K = [H2][I2]/[HI] must be set up using these equilibrium concentrations, allowing for the calculation of x. Solving this equation will yield the equilibrium concentration of H2.
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1. I am having trouble figuring out how to work the most basic equilibrium problem when there is an initial concentration of reactants only and no product. How am I supposed to find the constant with a zero in my quotient? Here is an example problem:
At 527˚C, K = 0.0160 for the following reaction: 2HI(g) ↔ H2 (g) + I2 (g)
What is the equilibrium concentration of H2
if the initial concentrations of both H2 and I2
= 0.0160 M and the initial concentration of
HI = 0 M?


2. k = product/reactant

3. If I knew where to start I wouldn't need help
 
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Bad-Wolf said:
1. I am having trouble figuring out how to work the most basic equilibrium problem when there is an initial concentration of reactants only and no product. How am I supposed to find the constant with a zero in my quotient? Here is an example problem:
At 527˚C, K = 0.0160 for the following reaction: 2HI(g) ↔ H2 (g) + I2 (g)
What is the equilibrium concentration of H2
if the initial concentrations of both H2 and I2
= 0.0160 M and the initial concentration of
HI = 0 M?


2. k = product/reactant

3. If I knew where to start I wouldn't need help

You mentioned that K=[product]/[reactants] - write this equation out in full detail and remember that they refer to the equilibrium values.
 
one way of solving this problem is by breaking it into parts.

initially,

= [I2] = 0.0160 mol/dm3
[HI] = 0 mol/dm3

at equilibrium, let [HI] be x mol/dm3
then, accordingly,

= [I2] = {0.0160 - (x/2)} mol/dm3, because 2 mol HI are produced by using 1 mol H2 and 1 mol I2.

find the equilibrium constant in term of x.
and then solve for x by equation the above expression to 0.0160

 
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