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3 years ago

Write the equilibrium constant formula for the following reaction: 2H2+O2 yields 2H2O

Chemistry

2 answers:

Anvisha [2.4K]3 years ago

8 0

$2 H_{2} + O_{2} ---\ \textgreater \ 2H_{2}O

$

$K= \frac{ [H_{2}O]^{2} }{ [H_{2}]^{2} [O_{2}]}$

Llana [10]3 years ago

4 0

Answer:

The expression for an equilibrium constant for the given reaction:

$K_c=\frac{[H_2O]^2}{[H_2]^2[O_2}}$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$2H_2+O_2\rightleftharpoons 2H_2O$

The expression for an equilibrium constant will be given by:

$K_c=\frac{[H_2O]^2}{[H_2]^2[O_2}}$

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3 years ago

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What is the pH of a 900 mL solution containing 3.40 grams of hydrocyanic acid

Allushta [10]

Answer:

$pH = 5.05$

Explanation:

pH is derived from the concentration of hydronium ions in a solution. Hydrocyanic acid is HCN.

First, we shall figure out the moles of HCN:

$\frac{3.4g}{27.03g/mol} = 0.125786$

If HCN was a strong acid:

HCN has a 1:1 ratio of H+ ions, the moles of H+ is also the same.

To find the molarity, we now divide by Liters. This gets us:

$\frac{0.125786 moles}{0.9L} = 0.139762 M$

Finally, we plug it into the definition of pH:

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

$pH = -log(0.139762)$

$pH = 0.855$

However, since HCN is a weak acid, it only partially dissociates. The $K_a$ of HCN is $6.2 * 10^{-10}$ .

$K_a = \frac{[H^+][A^-]}{[HA]}$

We can use an ice table to determine that when x = H+,

$K_a = \frac{x^2}{0.125786-x}$

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

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4 years ago

1. The Great Pacific Garbage Patch is the world's largest garbage dump. How big is it?

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Answer:

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3 years ago

3.364 g of hydrated barium chloride of BaCL2.xH2O was dissolved in water and made up to a total volume of 250.0 mL. 10.00 mL of

atroni [7]

Given:

Mass of hydrated barium chloride = 3.364 g

Total volume of barium chloride V(total)= 250 ml

Volume taken for titration V = 10 ml

Volume of AgNO3 consumed = 46.92 ml

Concentration of AgNO3 = 0.0253 M

To determine:

The value of x i.e. the water of hydration in BaCl2

Explanation:

The net ionic equation is-

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Based on the reaction stoichiometry: Equal moles of Ag+ and Cl- combine to form AgCl

Moles of Ag+ consumed = moles of Cl- present

Moles of Ag+ = V(AgNO3) * M(AgNO3) = 0.04692 * 0.0253 = 0.00119moles

Moles of Cl- present = 0.00119 moles

Thus, 0.00119 moles of Cl- are present in 10 ml of the solution

Therefore, number of moles of Cl- in 250 ml would be-

= 0.00119 * 250 /10 = 0.02975 moles of cl-

Now:

2 moles of Cl- are present in 1 mole of BaCl2

Therefore, 0.02975 moles of Cl- correspond to- 0.02975 * 1/2 = 0.01488 moles of BaCl2

Molar mass of BaCl2 = 208.22 g/mol

Thus, mass of BaCl2 = 0.01488 moles * 208.22 g.mol-1 = 3.098 g

Mass of water of hydration = 3.364 - 3.098 = 0.266 g

# moles of water 'x' = .266/18 = 0.015 ≅ 1

Ans: Formula for hydrated barium chloride = BaCl2. 1H2O

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4 years ago

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