Respuesta :
The given question is incomplete. But the complete question is this:
A mixture of [tex]CH_{4}[/tex] and [tex]H_2O[/tex] is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of [tex]H_2[/tex], 43.0 g of [tex]CH_{4}[/tex], and 48.4 g of [tex]H_{2}O[/tex]. Assuming that equilibrium has been reached, calculate [tex]K_{p}[/tex] for the reaction.
Explanation:
As the given reaction is as follows.
[tex]CH_4 + H_2O \rightarrow CO + 3H_2[/tex]
And, we know that
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
Therefore, calculate the moles as follows.
Moles of [tex]CH_4[/tex] = [tex]\frac{43}{16.04}[/tex]
= 2.6808 mol
Moles of [tex]H_2O[/tex] = [tex]\frac{48.4}{18.01528}[/tex]
= 2.6866 mol
Moles of CO = [tex]\frac{8.62}{28.01}[/tex]
= 0.307747 mol
Moles of [tex]H_{2}[/tex] = [tex]\frac{2.6}{2.01588}[/tex]
= 1.2897 mol
As, we know that
Concentration = [tex]\frac{moles}{volume (L) }[/tex]
Given volume = 5 L
Hence, calculate the concentration of given species as follows.
Conc. of [tex]CH_4 = \frac{2.6875}{5}[/tex]
= 0.5361
Conc. of [tex]H_2O = \frac{2.6889}{5}[/tex]
= 0.5373
Conc. of CO = [tex]\frac{0.307747}{5}[/tex]
= 0.06155
and, Conc. of [tex]H_2 = \frac{1.2897}{5}[/tex]
= 0.2579
Now, expression for equilibrium constant for the given reaction is as follows.
[tex]K_{c} = \frac{[CO][H_2]^{3}}{[CH_4][H_2O] }[/tex]
Now, putting the given values into the above formula as follows.
[tex]K_{c} = \frac{[0.06155][0.2579]^{3}}{[0.5361][0.5373]}[/tex]
[tex]K_{c} = 3.665 \times 10^{-3}[/tex]
Also, we know that
[tex]K_p = K_c \times (RT)^dn[/tex]
Consider the equation
[tex]CH_4(g) + H2O(g) \rightarrow CO(g) + 3H_2(g)[/tex]
Calculate change in moles of gas as follows.
change in gas moles (dn) = 1 + 3 - 1 - 1
dn = 2
As, [tex]K_p = K_c \times (RT)^2[/tex]
It is given that,
T = 1000 K , R = 0.0821
So,
[tex]K_p = 3.665 \times 10^{-3} \times (0.0821 \times 1000)^{2}[/tex]
[tex]K_p[/tex] = 24.70
Thus, we can conclude that value of [tex]K_{p}[/tex] for the reaction is 24.70.
