The enthalpy for the reaction : ΔH = -132
<h3>Further explanation</h3>
Given
Reaction and the enthalpy
Required
the enthalpy
Solution
Hess Law
Reaction 1 reverse :
A + B = G + C ΔH = -277
Reactions 2 and 3 remain the same (unchanged)
C + F = A ΔH = 303
D = B + H ΔH = -158
Add up all the reactions and remove the same compound from two different sides
D + F = G + H ΔH = -132
Answer:
<h2>Organic chemistry is the chemistry of carbon based compounds while biochemistry is the study of directly biologically relevant chemistry..</h2><h2>pls mark as brainliest</h2>
Answer:
they all have two electrons in their outermost shell
Answer:
-3135.47 kJ/mol
Explanation:
Step 1: Write the balanced equation
C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)
Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)
We will use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpies of formation
p: products
r: reactants
ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))
ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol
ΔH°r = -3135.47 kJ
Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.
