An equilibrium mixture of the following reaction was found to have [H2] = 0.020 M and [I2] = 0.020 M at 340°C. What is the concentration of HI?
H2(g) + I2(g)--> 2 HI(g)
Keq = 69 at 340°C
I got .166 but it said check number of sig figs so im guessing its 2 sig figs but I want to make sure:)
Thanks
two sig figures. .020 is two sig figures.
Thanks:)
To find the concentration of HI, you can use the equation for the equilibrium constant (Keq) and the given concentrations of H2 and I2.
The equation is: Keq = [HI]^2 / ([H2] * [I2])
Plug in the known values:
Keq = 69
[H2] = 0.020 M
[I2] = 0.020 M
Rearrange the equation to solve for [HI]:
[HI]^2 = Keq * ([H2] * [I2])
[HI]^2 = 69 * (0.020 * 0.020)
[HI]^2 = 0.0276
Taking the square root of both sides gives:
[HI] = √0.0276
Calculating the square root, the concentration of HI is approximately 0.166 M.
According to your calculations, you obtained 0.166 M, which seems to be correct. However, regarding the significant figures, you are right to consider them. Since both initial concentrations have two significant figures, the answer should also be rounded to two significant figures. Therefore, the final concentration of HI should be reported as 0.17 M.
So, the concentration of HI is 0.17 M (rounded to two significant figures).