calculate the change in enthalpy for Reaction 2.
Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Reaction 2: 3C3H8(g)+15O2(g)→9CO2(g)+12H2O(g), ΔH2=?

Best Answer: Sorry its kind of hard to explain, you just need to study hesslaw, practice makes perfect, keep studying!
A. REACTION 2 IS MULTIPLIED BY 3 (HESS LAW)
ΔH2= -6129 j
B. IF REACTION IS REVERSED YOU FLIP THE SIGN OF ΔH
ΔH = 92.3 kJ
C. KIND OF HARD TO EXPLAIN, BUT I CAN TRY AND SHOW YOU
EQUATION 1 (FLIP) CH4 --> C + H2
EQUATION 4 (FLIP AND MULTIPLY BY 2) H2O(g) --> H2O(l)
ΔH = -890.7

Answer Link

oleksandr130120 oleksandr130120 · Answer 2 · 2018-07-31T20:09:49+02:00

oleksandr130120 oleksandr130120

31-07-2018

[3 mol CO2(g) + 4 mol H2O(g)] - [1 mol C3H8(g) + 5 mol O2(g)]

[3 · (213.74 J/mol·K) + 4 · (188.825 J/mol·K)] - [1 · (269.9 J/mol·K) + 5 · (205.138 J/mol·K)]

=100.93 J/mol·K

Answer: 100.93 J/mol·K

Answer Link

calculate the change in enthalpy for Reaction 2. Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ Reaction 2: 3C3H8(g)+15O2(g)→9CO2(g)+12H2O(g), ΔH2=?

Respuesta :

Otras preguntas

calculate the change in enthalpy for Reaction 2.
Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ
Reaction 2: 3C3H8(g)+15O2(g)→9CO2(g)+12H2O(g), ΔH2=?