[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26
Answer:
r = 3.61x
M/s
Explanation:
The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.
r = k.![[S2O2^{-8} ]^{x} x [I^{-} ]^{y}](https://tex.z-dn.net/?f=%5BS2O2%5E%7B-8%7D%20%5D%5E%7Bx%7D%20x%20%5BI%5E%7B-%7D%20%5D%5E%7By%7D)
K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :
![\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}](https://tex.z-dn.net/?f=%5Cfrac%7Br1%7D%7Br2%7D%20%3D%20%5Cfrac%7B0.018%5E%7Bx%7D%20x0.036%5E%7By%7D%20%7D%7B0.027%5Exx0.036%5Ey%7D)
![\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y](https://tex.z-dn.net/?f=%5Cfrac%7B2.6x10%5E%7B-6%7D%7D%7B3.9x10%5E%7B-6%7D%7D%20%3D%20%28%5Cfrac%7B0.018%7D%7B0.027%7D%29%5Exx%28%5Cfrac%7B0.036%7D%7B0.036%7D%29%5Ey)
![0.67 = 0.67^x](https://tex.z-dn.net/?f=0.67%20%3D%200.67%5Ex)
x = 1
Now, to find the coefficient y let's do the same for the experiments 1 and 3:
![\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}](https://tex.z-dn.net/?f=%5Cfrac%7Br1%7D%7Br3%7D%20%3D%20%5Cfrac%7B0.018x0.036%5Ey%7D%7B0.036x0.054%5Ey%7D)
![\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y](https://tex.z-dn.net/?f=%5Cfrac%7B2.6x10%5E%7B-6%7D%7D%7B7.8x10%5E%7B-6%7D%7D%20%3D%20%28%5Cfrac%7B0.018%7D%7B0.036%7D%29x%28%5Cfrac%7B0.036%7D%7B0.054%7D%29%5Ey)
![0.33 = 0.5x 0.67^y](https://tex.z-dn.net/?f=0.33%20%3D%200.5x%200.67%5Ey)
![0.67 = 0.67^y](https://tex.z-dn.net/?f=0.67%20%3D%200.67%5Ey)
y = 1
Now, we need to calculate the constant k in whatever experiment. Using the first :
![2.6x10^{-6} = kx0.018x0.036](https://tex.z-dn.net/?f=2.6x10%5E%7B-6%7D%20%3D%20kx0.018x0.036)
![kx6.48x10^{-4} = 2.6x10^{-6}](https://tex.z-dn.net/?f=kx6.48x10%5E%7B-4%7D%20%3D%202.6x10%5E%7B-6%7D)
k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]
Using the data given,
r = ![4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}](https://tex.z-dn.net/?f=4.01x10%5E%7B-3%7Dx1.8x10%5E%7B-2%7Dx5.0x10%5E%7B-2%7D)
r = 3.61x
M/s
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wtjfavyw
Answer:
Yes
Explanation:
Masses for the three subatomic particles can be expressed in amu (atomic mass units) or grams. For simplicity, we will use the amu unit for the three subatomics. Both neutrons and protons are assigned as having masses of 1 amu each.