Question

1320 the reactions, 0 AH = -968.1 _XO XOS) CO. (9) AH +195,9 what is AH for this reaction? X65) +0,)+CO,CE) —- XCO, (3) N

Answer 1

Answer -

Given,

X (s) + 1/2 O₂ (g)  $\rightarrow$ XO (g)   $\Delta$H = -968.1 kJ

XCO₃ (s)  $\rightarrow$ XO (s) + CO₂ (g) $\Delta$H = +195.9 kJ

$\Delta$H of reaction

X (s) + 1/2 O₂ (g) + CO₂ (g)  $\rightarrow$ XCO₃ (s)

We know that,

X (s) + 1/2 O₂ (g)  $\rightarrow$ XO (g)   $\Delta$H = -968.1 kJ

XCO₃ (s)  $\rightarrow$ XO (s) + CO₂ (g) $\Delta$H = +195.9 kJ

Reverse of second reaction is

XO (s) + 1/2 O₂ (g) + CO₂ (g) $\rightarrow$ XCO₃ (s)  $\Delta$H = - 195.9 kJ [sign will be changed]

If we reverse the second reaction and add it to the first reaction, we will get the final reaction.

X (s) + 1/2 O₂ (g) + XO (s) + CO₂ (g) $\rightarrow$ XO (g) + XCO₃ (s)

XO (s) + 1/2 O₂ (g) + CO₂ (g) $\rightarrow$ XCO₃ (s)

So, $\Delta$H = $\Delta$H of reaction 1 + $\Delta$H of reverse reaction 2

$\Delta$H =  -968.1 kJ + (- 195.9 kJ )

$\Delta$H = - 1164 kJ [Answer]