Hess\'s Law A Find the AH for the reaction below, given the following reactions
ID: 503394 • Letter: H
Question
Hess's Law A Find the AH for the reaction below, given the following reactions and subsequent Delta H values: PCl_5(g) rightarrow PCl_3(g) + Cl_2(g) P_4(s) + 6Cl_2(g) rightarrow 4PCl_3(g) Delta H = -2439 kJ 4PCl_5(g) rightarrow P_4 (s) + 6l_2(g) Delta H = 3438 kJ Find the Delta H for the reaction below, given the following reactions and subsequent Delta H values: 2CO_2 (g) + H_2O(g) rightarrow C_2H_2(g) + 5/2O_2(g) C_2H_2(g) + 2H_2(g) rightarrow C_2H_6(g) Delta H = - 94.5 kJ H_2O (g) rightarrow H_2(g) + 1/2O_2 (g) Delta H = 71.2 kJ C_2H_6(g) + 7/20O_2(g) rightarrow 2CO_2 (g) + 3H_2O(g) Delta H = - 283 kJ Find the Delta H for the reaction below, given the following reactions and subsequent Delta H values: N_2H_4(I) + H_2 (g) rightarrow 2NH_3(g)
Explanation / Answer
Q1.
Apply hess law, so
from the reactions, we need PCl5 in the LEFT side, with 1 coefficient so:
divide equation 2, by 4
PCl5 = 1/4P4 + 10/4Cl H = 3438/4
PCl5 = 1/4P4 + 5/2Cl H = 859.5
for equation 1, we need PCl4 in the right side, so we must leave it as it is, divide by 4,
1/4P + 6/4Cl2 = PCl3 H = -2439/4 = 609.75
add all
1/4P + 3/2Cl2 + PCl5 = 1/4P4 + 5/2Cl +PCl3 H = 859.5+609.75
cancel common terms
PCl5 = PCl3 + Cl2 H = 1469.25 kJ/mol
Q2.
Apply same logic
now...
C2H2(g) must be in the right side, invert (RXN1)
C2H6 = C2H2 + 2H2 H = 94.5
we need to cancel H2, so invert and multipl RXN2 by 2
2H2 + O2 = 2H2O H = -2*(71.2)= - 142.4
finally, we need to cancel C2H6 from RXN1, we must invert (3)
2CO2 + 3H2O <--> C2H6 + 7/2O2 H = 283
Add all
C2H6 + 2H2 + O2= C2H2 + 2H2 +2H2O + C2H6 + 7/2O2 H = 94.5+ - 142.4+283
cancel common terms
2CO2 + 3H2O + C2H6 + 2H2 + O2= C2H2 + 2H2 +2H2O + C2H6 + 7/2O2 H = 94.5+ - 142.4+283
2CO2 + H2O = C2H2 + 5/2O H = 235.1 kJ