Chemistry, 02.07.2019 12:30 kait93

Asystem delivers 225 j of heat to the surroundings while delivering 645 j if work calculate the change in the internal chang

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Answer from: ozzie26

The change in the internal energy of the system is -870 J.

The given parameters;

heat added to the system, Q = 225 Jwork done on the surroundings, W = 645 J

The change in the internal energy of the system is calculated by applying the first law of thermodynamic as shown below;

ΔU = Q - W

where;

ΔU is the change in the internal energy of the systemQ is the heat addedW is the work done on the surrounding

Substitute the given parameters and solve for change in internal energy;

ΔU = (-225) - 645 J

ΔU = -870 J

Thus, the change in the internal energy of the system is -870 J.

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Answer from: brittneyzenger001

The change in the internal energy:-870 J

Further explanation

The laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created

The equation is:

$\rm \Delta U=Q+W$

Energy owned by the system is expressed as internal energy (U)

This internal energy can change if it absorbs heat Q (U> 0), or releases heat (U <0). Or the internal energy can change if the system does work or accepts work (W)

The sign rules for heat and work are set as follows:

• The system receives heat, Q +

• The system releases heat, Q -

• The system does work, W -

• the system accepts work, W +

Heat can be calculated using the formula:

Q = mc∆T

A 225 Delivers System J Of Heat To The Surroundings While Delivering 645 J Of Work, then

Q release heat = -225 J

The system does work, W = -645 J

the change in the internal energy, ∆E, of the system

∆E / ∆U = Q + W

∆E / ∆U = -225 - 645

∆E / ∆U = -870 J

Learn more

the difference between temperature and heat

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Specific heat

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relationships among temperature, heat, and thermal energy.

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When heat is added to a substance

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Answer from: Quest

Density because it’s holyyy

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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