Consider the following processes: ΔH (kJ/mol) 1/2A → B +150 3B → 2C + D E + A → 2D + 350 For B + D → E + 2C, ΔH will be:

The processes are as shown below.
ΔH
1/2A → B +150 (1)
3B → 2C + D (2)
E + A → 2D + 350 (3)
________________________
B + D → E + 2C (4)
From equation (1), 1/2A → B; ΔH = +150 kJ/mol
Therefore, A → 2B; ΔH = +300 kJ/mol (1a)
From equation (2), 3B → 2C + D; ΔH = - kJ/mol
From equation (3), E + A → 2D + 350 kJ/mol
Equation (4) can be obtained by adding (1a) + (2) - (3) :
A → 2B; ΔH = +300 kJ/mol (1a)
3B → 2C + D; ΔH = - kJ/mol
E + A → 2D + 350 kJ/mol (3)

2B + 3B + E + A → 2B + 2C + D + 2D + 350 kJ/mol
5B + E + A → 2C + 3D + 350 kJ/mol
For equation (4):
B + D → E + 2C
ΔH4 = ΔH1a + ΔH2 – ΔH3
ΔH4 = 300 + x - 350
ΔH4 = x-50 kJ/mol
Let's use another approach.
Reverse equation (1): B → ½A; ΔH = -150
Reverse equation (3): 2D → E + A; ΔH = -350
Add equation (2) and the above:
3B + 2D → 2C + D + E + A
3B + D → 2C + E + A (5)
Multiply equation (1) by 2 : A → 2B; ΔH = 300
Add equation (5) and the above:
A + 3B + D → 2C + E + A + 2B
3B + D → 2C + E + 2B
Equation (4): B + D → E + 2C
Let’s solve this using Hess’s law. Consider the following:
1/2A → B; ΔH = +150 kJ/mol (1)
3B → 2C + D; ΔH = x kJ/mol (2)
E + A → 2D + 350 kJ/mol (3)
B + D → E + 2C; ΔH = y kJ/mol (4)
Add (1) x 2 and (2) to get 7B → 2C + D + 2B; ΔH = 300 + x
Reverse (3) to get E + A → 2D + 350; ΔH = -350
(300 + x) + (-350) = y
This simplifies to x - 50 = y
There is insufficient information to solve for x and y.