# Question & Answer: Given the following equations: A (g) → B (g)   ΔH = –50 kJ…..

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:

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