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Given the following equations:
A (g) → B (g) ΔH = –50 kJ
B (g) → C (g) ΔH = –100 kJ
Calculate the enthalpy changes for the following.
Enter your answer in kJ but don’t include units. For example, if the answer is 35 kJ, you would enter “35” without the quotes.
2 A (g) → 2 B (g) has an enthalpy change of:
, C (g) → B (g) has an enthalpy change of:
, A (g) → C (g) has an enthalpy change of:
, 2 B (g) → A (g) + C (g) has an enthalpy change of:
Expert Answer
This requires Hess Laws, which implies that we can modify the coefficients of the reacitons as well as the values of HRxn
then..
a)
for 2A = 2B
we have only A = B,
we notice a pattern, it has 2x
therefore
HRxn of 2A = 2B = 2x ( HRxn of A = B)
substitute
HRxn of 2A = 2B = 2x ( -50 kJ) = -100 kJ
then
HRxn = -100 kJ
b)
for C = B, we notice it is the REVERSE reaction, therefore, dH = -dH
so
Hnew = -Horiginal = -(1)(-100) = +100 kJ
c)
A = C this requires much more modification
A = B and B = C can be modified
so
A +B = B+ C ; b cancells out
A = C
So
HRxn total = H1 + H2 = -50 + -100 = -150 kJ
d)
this is simple:
reaction 1 must be intverted
B = A H = -1*(-50) = + 50
B = C H = -100
add all
2B = A + C Htotal = +50 – 100 = -50 kJ
