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Measurements show that enthalpy of a mixture of gaseous reactants decreases by 228. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -55kJ of work is done on the mixture during the reaction.

Calculate the change in energy of the gas mixture during the reaction.

Is the reaction exothermic or endothermic?

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Answer:

-283 KJ, exothermic

Explanation:

We are given that

Enthalpy of a mixture of a gases reactants

Change in enthalpy=[tex]\Delat H[/tex]=-228 KJ

Where negative sign represents the enthalpy decreases.

Pressure=Constant.

Work don=w=-55 KJ

We have to calculate the change in energy of the gas mixture during the reaction.

At constant pressure, the change in enthalpy

[tex]\Delta H=\Delta U+P\Delta V[/tex]

Where w=[tex]-P\Delta V[/tex]

Where P=Constant

[tex]\Delta V=[/tex] Change in volume

[tex]\Delta U=[/tex] Change in energy

[tex]\Delta U=\Delta H-P\Delta V[/tex]

[tex]\Delta U=\Delta H+w[/tex]

Substitute the values then we get

[tex]\Delta U=-228-55=-283[/tex] KJ

Hence, the change in energy of the gas mixture during the reaction=-283 KJ

Change in enthalpy is negative it means the reaction is exothermic because the energy is evolved.

pstnonsonjoku

The reaction is exothermic, heat is evolved.

From the first law of thermodynamics we know that energy is neither created nor destroyed.

Therefore, the change in energy is obtained from the formula;

ΔU = ΔH + w

ΔU = change in energy

ΔH = Change in enthalpy (heat absorbed or evolved)

w = work done

ΔU = ?

w =  -55kJ

ΔH = -228. kJ

ΔU= -228. kJ + ( -55kJ)

ΔU = -283 kJ

Learn more: https://brainly.com/question/4147359

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