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3 years ago

13

Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the sta

rting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 200.mL flask with 2.4 atm of methane gas and 3.9 atm of water vapor at 46.0°C. She then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 6.5 atm. Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture.

1 answer:

Hatshy [7]3 years ago

5 0

<u>Answer:</u> The pressure equilibrium constant for the reaction is 1473.8

<u>Explanation:</u>

We are given:

Initial partial pressure of methane gas = 2.4 atm

Initial partial pressure of water vapor = 3.9 atm

Equilibrium partial pressure of hydrogen gas = 6.5 atm

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

$CH_4+H_2O\rightleftharpoons CO+3H_2$

<u>Initial:</u> 2.4 3.9

<u>At eqllm:</u> 2.4-x 3.9-x x 3x

Evaluating the value of 'x':

$\Rightarrow 3x=6.5\\\\x=2.167$

So, equilibrium partial pressure of methane gas = (2.4 - x) = [2.4 - 2.167] = 0.233 atm

Equilibrium partial pressure of water vapor = (3.9 - x) = [3.9 - 2.167] = 1.733 atm

Equilibrium partial pressure of carbon monoxide gas = x = 2.167 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:

$K_p=\frac{2.167\times (6.5)^3}{0.233\times 1.733}\\\\K_p=1473.8$

Hence, the pressure equilibrium constant for the reaction is 1473.8

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2 years ago

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2 years ago

How many atoms are in 8.28 moles of aluminum?

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Answer:

49.86 × 10²³ atoms of Al

Explanation:

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Number of moles of Al = 8.28 mol

Number of atoms = ?

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The given problem will solve by using Avogadro number.

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The number 6.022 × 10²³ is called Avogadro number.

For example,

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2 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

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2 years ago