Answer to Question #266975 in General Chemistry for kaltook

Question #266975

How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 18.5 L of H2 at 75°C and 744 torr?


1
Expert's answer
2021-11-17T15:27:03-0500

CO+2H2CH3OHCO+2H_2\to CH_3OH

VCO=15L=0.015m3V_{CO}=15L=0.015m^3

P=112000PaP=112000Pa

T=850C=358,15KT=85^0C=358,15K

PV=nRTPV = nR T

n=(PV)/(RT)n=(PV)/(RT)

n=(1120000,015)/(8,314358,15)n= (1120000,015)/(8,314358,15)

n(CO)=0,564mol

VCO=18,5L=0,0185m3V_{CO}=18,5L=0,0185m^3

P=744torr=99191,84PaP=744torr=99191,84Pa

T=750C=348,15KT=75^0C =348,15K

PV=nRTPV =n RT

n=(PV)/(RT)n=(PV)/(RT)

n=(99191,840,0185)/(8,314348,15)n=(99191,840,0185)/(8,314348,15)

n(H2)=0,634moln(H_2)=0,634mol

It can be seen that H2H_2 is the limiting reactant

n(CH3OH)=1/2n(H2)=1/20,634MOL=0,317moln(CH_3OH)=1/2n(H_2)=1/20,634MOL=0,317mol

ΔHrxn=ΔHf(CH3OH)ΔHf(CO)=238,6(110,5)=128,1kJ/mol\Delta H_{rxn}=\Delta H_f(CH_3OH)-\Delta H_f(CO)=-238,6-(-110,5)=-128,1kJ/mol

E=ΔHrxn×n=40,6kJE=\Delta H_{rxn}\times n =-40,6kJ


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