Question #0d7fc

1 Answer
Nov 20, 2016

#Δ_"rxn"H = "-1900 kJ, -930 kJ, -1900 kJ, and +933 kJ/mol CF"_4#, respectively.

Explanation:

#"C(s) + 2F"_2 "(g) → CF"_4 "(g)"; Δ_text(rxn)H = "-933 kJ/mol CF"_4#

(a) Using 2 mol of #"C"#

#Δ_text(rxn)H = 2.0 color(red)(cancel(color(black)("mol C"))) × (1 color(red)(cancel(color(black)("mol CF"_4))))/(1 color(red)(cancel(color(black)("mol C")))) × "-933 kJ"/(1 color(red)(cancel(color(black)("mol CF"_4)))) = "-1900 kJ"#

(b) Using 2 mol of #"F"_2#

#Δ_text(rxn)H = 2.0color(red)(cancel(color(black)("mol F"_2))) × (1 color(red)(cancel(color(black)("mol CF"_4))))/(2 color(red)(cancel(color(black)("mol F"_2)))) × "-933 kJ"/(1 color(red)(cancel(color(black)("mol CF"_4)))) = "-930 kJ"#

(c) Using 2 mol of #"CF"_4#

#Δ_text(rxn)H = 2.0 color(red)(cancel(color(black)("mol CF"_4))) × "-933 kJ"/(1 color(red)(cancel(color(black)("mol CF"_4)))) = "-1900 kJ"#

(d) Reversing the reaction

#"CF"_4"(g)" → "C(s) + 2F"_2"(g)"; Δ_text(rxn)H = "+933 kJ/mol CF"_4#