Question #99693

Given the thermochemical equations

X

2

+

3

Y

2

⟶

2

XY

3

Δ

H

1

=

−

340

kJ

X

2

+

2

Z

2

⟶

2

XZ

2

Δ

H

2

=

−

140

kJ

2

Y

2

+

Z

2

⟶

2

Y

2

Z

Δ

H

3

=

−

240

kJ

Calculate the change in enthalpy for the reaction.

4

XY

3

+

7

Z

2

⟶

6

Y

2

Z

+

4

XZ

2

X

2

+

3

Y

2

⟶

2

XY

3

Δ

H

1

=

−

340

kJ

X

2

+

2

Z

2

⟶

2

XZ

2

Δ

H

2

=

−

140

kJ

2

Y

2

+

Z

2

⟶

2

Y

2

Z

Δ

H

3

=

−

240

kJ

Calculate the change in enthalpy for the reaction.

4

XY

3

+

7

Z

2

⟶

6

Y

2

Z

+

4

XZ

2

Expert's answer

((2*2nd reaction) +(3*3rd reaction)) - (2*1st reaction) = The reaction in the problem

2X2+4Z2 --> 4XZ2

6Y2+3Z2 --> 6Y2Z

4XY3 --> 2X2+6Y2

-------------------------------

4XY3+ 7Z2--> 6Y2Z + 4XZ2

Therefore the change in enthalpy=

(-280 - 720 +680) kj = - 320 kj

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