Chemistry, 23.03.2020 21:09, darkskinnednune

Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 1.5 L flask with 0.60 atm of methane gas and 2.6 atm of water vapor at 47. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 1.4 atm. Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits x10.

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answered: noahalderman83

The pressure equilibrium constant for the reaction is $4.5\times 10^0$

Explanation:

We are given:

Initial partial pressure of methane = 0.60 atm

Initial partial pressure of water vapor = 2.6 atm

Equilibrium partial pressure of hydrogen gas = 1.4 atm

The chemical equation for the reaction of methane gas and water vapor follows:

$CH_4+H_2O\rightleftharpoons CO+3H_2$

Initial: 0.60 2.6

At eqllm: 0.60-x 2.6-x x 3x

Evaluating the value of 'x':

$\Rightarrow 3x=1.4\\\\x=0.467$

So, equilibrium partial pressure of methane gas = (0.60 - x) = [0.60 - 0.467] = 0.133 atm

Equilibrium partial pressure of water vapor = (2.6 - x) = [2.6 - 0.467] = 2.133 atm

Equilibrium partial pressure of carbon monoxide gas = x = 0.467 atm

The expression of K_p for above equation follows:

$K_p=\frac{p_{CO}\times (p_{H_2})^3}{p_{CH_4}\times p_{H_2O}}$

Putting values in above equation, we get:

$K_p=\frac{0.467\times (1.4)^3}{0.133\times 2.133}\\\\K_p=4.517\times 10^0$

Hence, the pressure equilibrium constant for the reaction is $4.5\times 10^0$

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

answered: Guest

That is a lot of moles that you have

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Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hy...