NH3(g)+HBr(g)→NH4Br(s)
CaCO3(s)→CaO(s)+CO2(g)
CH4(g)+3Cl2(g)→CHCl3(g)+3HCl(g)
ΔG∘f for CHCl3(g) is - 70.4 kJ/mol
Been stuck on this all day!
You need information that you haven't provided in the question. Perhaps your text has a table of ΔGof for these compounds. In that case,
ΔGorxn = Sum of ΔGof products - ΔGof reactants
Just plug in the values that you find and calculate the free energy change of the reaction.
ΔG = ΔG formation of products - ΔG formation of reactants
ΔG = ΔG∘f [ NH4Br(s) ] - [ ΔG∘f (NH3(g) + ΔG∘F (HBr(g) ]
Plug in the values ( check back of your textbook for ΔG∘formation values
ΔG = [ΔGf (CHCl3(g)) + 3 x ΔGf (HCl(g) ] - [ ΔGf (CH4(g) + 3x ΔGf (CL2(g)) ]
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Verified answer
You need information that you haven't provided in the question. Perhaps your text has a table of ΔGof for these compounds. In that case,
ΔGorxn = Sum of ΔGof products - ΔGof reactants
Just plug in the values that you find and calculate the free energy change of the reaction.
ΔG = ΔG formation of products - ΔG formation of reactants
ΔG = ΔG∘f [ NH4Br(s) ] - [ ΔG∘f (NH3(g) + ΔG∘F (HBr(g) ]
Plug in the values ( check back of your textbook for ΔG∘formation values
CH4(g)+3Cl2(g)→CHCl3(g)+3HCl(g)
ΔG = [ΔGf (CHCl3(g)) + 3 x ΔGf (HCl(g) ] - [ ΔGf (CH4(g) + 3x ΔGf (CL2(g)) ]