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2 years ago

What happens if more product is added to a system at equilibrium apex?

1 answer:

6 0

According to Le Chatelier's Principle," If a system (reaction) at equilibrium is subjected to any external stress, either by changing pressure, temperature or concentration of reactants or products, the system will adjust itself in such a way that the effect is minimized".

So, when more product is added to reaction at equilibrium, the equilibrium will shift in backward direction in order to minimize the effect of concentration of product.

Result:
With the addition of product, the rate of backward reaction will increase resulting in increasing the concentration of reactants.

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Name the alkanes please help

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4 0

1 year ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

2 years ago

So, a gas, and an

valina [46]

Answer:

d) cut the large sized Cu solid into smaller sized pieces

Explanation:

The aim of the question is to select the right condition for that would increases the rate of the reaction.

a) use a large sized piece of the solid Cu

This option is wrong. Reducing the surface area decreases the reaction rate.

b) lower the initial temperature below 25 °C for the liquid reactant, HNO3

Hugher temperatures leads to faster reactions hence this option is wrong.

c) use a 0.5 M HNO3 instead of 2.0 M HNO3

Higher concentration leads to increased rate of reaction. Hence this option is wrong.

d) cut the large sized Cu solid into smaller sized pieces

This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.

5 0

2 years ago

Hello,

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

7 0

2 years ago

Two important ways that energy is transported in the world around us is through

Ann [662]

Explanation:

The two ways that energy can be transferred are by doing work and by heat transfer.

5 0

2 years ago