Consider this equilibrium: I2(s) + H2O(l) H + (aq) + I-(aq) + HOI(aq). What is the Keq expression for this equilibrium?

1 answer:

3 0

<span>The Keq expression for I2(s) + H2O(l) H + (aq) + I-(aq) + HOI(aq) equilibrium is H2(g) + I2(g) <----> 2 HI(g).In this Keq expression hydrogen reacts with iodine gives Hydrogen Iodide.HI is a diatomic molecule and it is one of the primary source of iodine which act as a reducing agent.Totally it is the reverse chemical equilibrium reaction.</span>

You might be interested in

I don’t understand this question and need help!!

Dafna1 [17]

Medium about 3 second? Not sure lol just need more points honestly lol

6 0

2 years ago

When 2 moles of Fe(s) react with Cl2(g) to form FeCl3(s) according to the following equation, 799 kJ of energy are evolved.

GenaCL600 [577]

Answer:

1. Exothermic.

2. -1598 kJ.

Explanation:

Hello!

1. In this case, according to the reaction, we can infer that 799 kJ of energy are evolved (given off, released) it means that the enthalpy of reaction is negative as the reactants have more energy than the products; which means this is an exothermic reaction.

2. Here, as we know that the enthalpy of reaction is -799 kJ/mol, we can compute the q-value as shown below, considering the reacted 2 moles of solid iron:

$q=2mol*-799 kJ/mol\\\\q=-1598kJ$

Which means that 1598 kJ of energy are evolved when 2 moles of solid iron react.

Best regards!

4 0

3 years ago

What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent

dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

$\text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}$

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, $\text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}$