Chemistry, 24.12.2019 22:31 Carlos448

Amixture of ch4 and h2o 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 h2, 43.0 g of ch4, and 48.4 g of h2o.

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Answer from: lakaralake11

The given question is incomplete. But the complete question is this:

A mixture of $CH_{4}$ and H_2O 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 H_2 , 43.0 g of $CH_{4}$ , and 48.4 g of $H_{2}O$ . Assuming that equilibrium has been reached, calculate $K_{p}$ for the reaction.

Explanation:

As the given reaction is as follows.

$CH_4 + H_2O \rightarrow CO + 3H_2$

And, we know that

No. of moles = $\frac{mass}{\text{molar mass}}$

Therefore, calculate the moles as follows.

Moles of CH_4 = $\frac{43}{16.04}$

= 2.6808 mol

Moles of H_2O = $\frac{48.4}{18.01528}$

= 2.6866 mol

Moles of CO = $\frac{8.62}{28.01}$

= 0.307747 mol

Moles of $H_{2}$ = $\frac{2.6}{2.01588}$

= 1.2897 mol

As, we know that

Concentration = $\frac{moles}{volume (L)
}$

Given volume = 5 L

Hence, calculate the concentration of given species as follows.

Conc. of $CH_4 = \frac{2.6875}{5}$

= 0.5361

Conc. of $H_2O = \frac{2.6889}{5}$

= 0.5373

Conc. of CO = $\frac{0.307747}{5}$

= 0.06155

and, Conc. of $H_2 = \frac{1.2897}{5}$

= 0.2579

Now, expression for equilibrium constant for the given reaction is as follows.

$K_{c} = \frac{[CO][H_2]^{3}}{[CH_4][H_2O]
}$

Now, putting the given values into the above formula as follows.

$K_{c} = \frac{[0.06155][0.2579]^{3}}{[0.5361][0.5373]}$

$K_{c} = 3.665 \times 10^{-3}$

Also, we know that

$K_p = K_c \times (RT)^dn$

Consider the equation

$CH_4(g) + H2O(g) \rightarrow CO(g) + 3H_2(g)$

Calculate change in moles of gas as follows.

change in gas moles (dn) = 1 + 3 - 1 - 1

dn = 2

As, $K_p = K_c \times (RT)^2$

It is given that,

T = 1000 K , R = 0.0821

So,

$K_p = 3.665 \times 10^{-3} \times (0.0821 \times 1000)^{2}$

K_p = 24.70

Thus, we can conclude that value of $K_{p}$ for the reaction is 24.70.

Answer from: Quest

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Answer from: Quest

It is 4.92 e-4 sorry if i am wrong

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