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NeX [460]
3 years ago
11

If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be

Physics
2 answers:
nekit [7.7K]3 years ago
8 0
<h3><u>Answer;</u></h3>

<em><u>Large </u></em>

If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be <em><u>large.</u></em>

<h3><u>Explanation;</u></h3>
  • <em><u>Chemical equations may either be exothermic or endothermic based on the energy changes in the reaction. Exothermic reactions are those reactions that loose heat energy to the surrounding, therefore the enthalpy change in these reactions is negative.</u></em> Endothermic reaction on the other hand are those reactions in which energy is absorbed from the reaction thus their enthalpy is positive.
  • <em><u>Chemical equilibrium</u></em> in chemical reactions occurs when the rate of forward reaction is equal to the rate of reverse reaction. <em><u>Equilibrium constant is the value of the chemical reaction quotient when the equilibrium of a reaction is achieved.</u></em>
  • <em><u>If a reaction is exothermic and produces large quantities of heat in order to reach equilibrium then the equilibrium constant will be large.</u></em>
natima [27]3 years ago
5 0
The equilibrium constant will be lowered and the equilibrium will shift to the left if the heat being produced is not removed.
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Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\

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Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\

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As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of
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Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

\Delta T_f=K_f\times m

where,

\Delta T_f =depression in freezing point =  

K_f = freezing point constant  

m = molality  ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

  1. If molality of the solution in high the depression in freezing point of the solution will be more.
  2. If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

\frac{p_o-p_s}{p_o}=\chi_{solute}

p_o = Vapor pressure of pure solvent

p_s  = Vapor pressure of solution

\chi_{solute} = Mole fraction of solute

p_s\propto \frac{1}{\chi_{solute}}

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

  1. Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
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