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Rina8888 [55]
2 years ago
11

Given that this reaction is exothermic, what direction will the equilibrium shift when the temperature of the reaction is decrea

sed? 3H₂(g) + N₂(g) ⇔ 2NH₃(g)
A. Left (towards the reactants)
B. Right (towards the products)
C. No Shift
Chemistry
1 answer:
Svetradugi [14.3K]2 years ago
7 0

Equilibrium will shift towards the products when temperature is decreased in an exothermic reaction of the  formation of ammonia.

<h3>What is an exothermic reaction?</h3>

An exothermic reaction is a reaction in which heat content of the reactants is greater than the heat content of product.

In an exothermic reaction, heat is given off.

For an exothermic reaction in equilibrium, increasing temperature shifts equilibrium to the towards the left, towards the reactants.

On the other, equilibrium will shift towards the products when temperature is decreased.

Therefore, equilibrium will shift towards the products when temperature is decreased in the reaction of the  formation of ammonia.

Learn more about exothermic reactions at: brainly.com/question/13892884

#SPJ1

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formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b
HACTEHA [7]

Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

3 0
3 years ago
A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final
VMariaS [17]
M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C

m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C

T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
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