Chemistry, 04.11.2019 15:31 deehunchoo

Reaction 1: c3h8(g)+5o2(g)→3co2(g)+4h2o(g), δh1 = −2043 kj.
reaction 2: 3c3h8(g)+15o2(g)→9co2(g)+12h2o(g), δh2 = ?

determine the value of δh2.

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Answer from: KATE1223

From the given balanced chemical reactions, it may be observed that to obtain the second one, we just to multiply all the numerical coefficients of the first by 3. This implies that the value of ΔH2 is also thrice that of ΔH1. Therefore, the value of ΔH2 is -6129 kJ.

Answer from: Quest

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there are, however, a multitude of problems related to this immediate, push-back reaction of anger. and probably the key one is that almost never does it resolve the issue that gave rise to it. such reactive anger is probably best understood as self-defeating. as david burns, the author of the seminal self- book feeling good, observes: if realistically, acting on your anger is to make any sense, it needs to meet two criteria—which, in almost every case, is frankly impossible. that is, your anger must:

be directed toward a person who has intentionally (and needlessly) behaved in a hurtful way toward you; and

be beneficial or advantageous to you (i.e., assist you in achieving a desired goal).

i think you’ll agree that only rarely can you claim that your anger is both warranted and , whether to yourself or the relationship.

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explanation:

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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Reaction 1: c3h8(g)+5o2(g)→3co2(g)+4h2o(g), δh1 = −2043 kj.
reaction 2: 3c3h8(g)+15o2(g)→9c...

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