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ratelena [41]
3 years ago
5

Consider the reaction 2 SO2(g) + O2(g) <=> 2 SO3(g), which is exothermic as written. What would be the effect on the equil

ibrium position of removing O2(g)?
Chemistry
1 answer:
enot [183]3 years ago
5 0

Answer:

Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.

Explanation:

The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.

The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.

Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.

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Comment Both propane and benzene are hydrocarbons. As a rule,
kozerog [31]

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

5 0
3 years ago
You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.
Nitella [24]

The question requires us to use the dilution formula M_iV_i=M_fV_f, where M_i and V_i are the stock concentration and volume respectively, then M_f and V_f are the dilute concentration and volume respectively.

a. C_s_t_o_c_k= 1.5 M KCI, C_d_i_l_u_t_e=0.20M KCl, V_d_i_l_u_t_e=250ml KCl

b.M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml

To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.


8 0
3 years ago
Express as ordinary numbers. 3 x 100 =
Stolb23 [73]
The answer is 300
3 * 100
3 * 10 * 10
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300

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7 0
3 years ago
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While heating two different samples, of
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A student performed an investigation at sea level. First she placed 400 mL of water in four different containers. Then she place
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The answer is b your welcome
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