Solving this question in a certain way

[nhrock3]

this is the general way i solve every question:
example of a solved question in this way

we put 0.8 mole of H2 1.3 mole of I2 in one litter cantainer

the balance constant is 10.7
calculate how much 2HI will be in the container when the reaction balances

H2 +I2 = 2HI
start 0.8 1.3 0
moves -x -x +2x
end 0.8-x 1.3-x 2x

there are two solutions but only one is possiblr
x=0.0604
[HI]=0.12how to solve this question in the same way
we mix 0.5 moles of A
1 mole of B
2 moles of C
in a one litter container

after it balances we find 0.6 moles of A
find K for
=(is a sign where two arrows go in different directions)
2A =B+3c

 

[chemisttree]

Very nice! Glad to be of assistance.

 

[Blueshift5]

First, we need to set up the balanced chemical equation for the reaction:
2A = B + 3C

Next, we can use the given initial amounts of each reactant and the balanced equation to set up a table similar to the one used in the previous example:

Reactant | Initial Amount | Change | Equilibrium Amount
A | 0.5 mol | -2x | 0.6 mol
B | 1 mol | -x | x mol
C | 2 mol | -3x | 3x mol

Using the equilibrium amounts, we can set up an expression for the equilibrium constant (K) using the formula: K = [products]/[reactants]:

K = ([C]^3)/[A]^2

Plugging in the equilibrium amounts from the table, we get:

K = (x * (3x)^3) / (0.6 - 2x)^2

To solve for x, we can use the quadratic formula since this equation cannot be solved algebraically:

x = 0.083 mol

Therefore, the equilibrium constant (K) for this reaction is:

K = (0.083 * (3*0.083)^3) / (0.6 - 2*0.083)^2
K = 0.00184

This is the general way to solve for the equilibrium constant using the initial and equilibrium amounts of the reactants in a given reaction.