The enthalpy changes for two reactions are given by the equations:
`2Cr(s) +(3)/(2)O_(2)(g) rarr Cr_(2)O_(3)(s), DeltaH^(Theta) =- 1130 kJ`
`C(s) +(1)/(2)O_(2)(g)rarr CO(g), DeltaH^(Theta) =- 110 kJ`
What is the enthalpy change in `kJ` for the following reactions?
`3C(s) +Cr_(2)O_(3)(s) rarr 2Cr(s) +3CO(g)`

Options

A. `-1460kJ`
B. `-800kJ`
C. `+800kJ`
D. `+1020kJ`

Answer

C

Explanation

The given equations can be written as:
`Cr_(2)O_(3)(s)to2Cr(s)+(3)/(2)O_(2)(g)," "DeltaH=+1130kJ`
`3C(s)+(3)/(2)O_(2)(g)to3CO(g)," "DeltaH=-330kJ`
Adding the above equations, we get:
`Cr_(2)O_(3)(s)+3C(s)to2Cr(s)+3CO(g)," "DeltaH=+800kJ`