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The figure below shows an energy cycle. a) Give the name of the enthalpy change H1. b) Determine the value of H2.

Chemistry

ScanSolve AI Answer

Kutokana na swali lililopita, fomula ya empiriki ya alkanol ilikuwa C4H10OC_4H_{10}O. Ikiwa fomula ya empiriki na fomula ya molekuli ni sawa, basi fomula ya molekuli pia ni C4H10OC_4H_{10}O. Hii inawakilisha butanol.

b) Muundo wa alkanol (butan-1-ol):

CH3−CH2−CH2−CH2−OHCH_3-CH_2-CH_2-CH_2-OH

Sasa, kwa swali la 3:

a) Jina la mabadiliko ya entalpia ΔH1\Delta H_1:
ΔH1\Delta H_1 inawakilisha uundaji wa maji ya gesi (H2O(g)H_2O_{(g)}) kutoka kwa elementi zake (H2(g)H_{2(g)} na O2(g)O_{2(g)}) katika hali zao za kawaida.
Jina lake ni Standard enthalpy of formation of gaseous water (Entalpia Sanifu ya Uundaji wa Maji ya Gesi).

b) Kuamua thamani ya ΔH3\Delta H_3:
Kutokana na Sheria ya Hess, mabadiliko ya jumla ya entalpia katika mzunguko ni sifuri. Tunaweza kuandika uhusiano kati ya entalpia hizi:

ΔH2=ΔH1+ΔH3\Delta H_2 = \Delta H_1 + \Delta H_3

Tunataka kupata ΔH3\Delta H_3, kwa hivyo tunapanga upya fomula:

ΔH3=ΔH2−ΔH1\Delta H_3 = \Delta H_2 - \Delta H_1

Tunapewa thamani zifuatazo:
ΔH1=−187.8 kJ mol−1\Delta H_1 = -187.8 \text{ kJ mol}^{-1}
ΔH2=−285.8 kJ mol−1\Delta H_2 = -285.8 \text{ kJ mol}^{-1}

Badilisha thamani hizi kwenye fomula:

ΔH3=(−285.8 kJ mol−1)−(−187.8 kJ mol−1)\Delta H_3 = (-285.8 \text{ kJ mol}^{-1}) - (-187.8 \text{ kJ mol}^{-1}) ΔH3=−285.8 kJ mol−1+187.8 kJ mol−1\Delta H_3 = -285.8 \text{ kJ mol}^{-1} + 187.8 \text{ kJ mol}^{-1} ΔH3=−98.0 kJ mol−1\Delta H_3 = -98.0 \text{ kJ mol}^{-1}

Thamani ya ΔH3\Delta H_3 ni: −98.0 kJ mol−1\boxed{-98.0 \text{ kJ mol}^{-1}}

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