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Salsk061 [2.6K]
4 years ago
14

Consider the following system at equilibrium:D(aq)+E(aq)<=>F(aq)Classify each of the following actions by whether it cause

s a leftward shift, a rightward shift, or no shift in the direction of the net reaction.Increase DIncrease EIncrease FDecrease DDecrease EDecrease FTriple D and reduce E to one thirdTriple both E and F
Chemistry
1 answer:
igomit [66]4 years ago
4 0

Explanation:

D(aq) + E(aq) <=> F(aq)

This question is based on Le Chatelier's principle.

Le Chatelier's principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

Increase D

D is a reactant. if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

Increase E

E is a reactant. if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

Increase F

F is a product.  If we add additional product to a system, the equilibrium will shift to the left, in order to produce more reactants. The reaction would shift to the left.

Decrease D

if we remove reactants from the system, equilibrium will  be shifted to the left.

Decrease E

if we remove reactants from the system, equilibrium will  be shifted to the left.

Decrease F

if we remove products from the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

Triple D and reduce E to one third

no shift in the direction of the net reaction, Both changes cancels each other.

Triple both E and F

no shift in the direction of the net reaction, Both changes cancels each other.

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Read 2 more answers
2C(s)+O2(g)2H2(g)+O2(g)H2O(l)→2CO(g)→2H2O(g)→H2O(g)ΔHΔHΔH=−222kJ=−484kJ=+44kJ Use the thermochemical data above to calculate the
sveticcg [70]

Answer : The change in enthalpy for the reaction is, 175 kJ

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According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The main reaction is:

H_2O(l)+C(s)\rightarrow CO(g)+H_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2C(s)+O_2(g)\rightarrow 2CO(g)     \Delta H_1=-222kJ

(2) 2H_2(g)+O_2(g)\rightarrow 2H_2O(g)    \Delta H_2=-484kJ

(3) H_2O(l)\rightarrow H_2O(g)    \Delta H_3=44kJ

Now we are dividing reaction 1 by 2, dividing reverse reaction 2 by 2 and then adding all the equations, we get :

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=\frac{-222kJ}{2}=-111kJ

(2) H_2O(g)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_2=\frac{484kJ}{2}=242kJ

(3) H_2O(l)\rightarrow H_2O(g)    \Delta H_3=44kJ

The expression for change in enthalpy of the given reaction is:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-111)+(242)+(44)

\Delta H=175kJ

Therefore, the change in enthalpy for the reaction is, 175 kJ

6 0
4 years ago
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