Calculate the standard enthalpy change, ΔH, for the formation of 1 mol of strontium carbonate (the...

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Calculate the standard enthalpy change, ΔH, for the formation of 1 mol of strontium carbonate (the...

Calculate the standard enthalpy change, ΔH, for the formation of 1 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements.

Sr(s) + C(graphite) + 32 O2(g) -----> SrCO3(s)

The information available is:

(1) Sr(s)+ 12O2(g) ---->SrO(s) DeltaH= -592kj

(2) SrO(s) + CO2(g) -----> SrCO3(s) DeltaH= -234kj

(3) C(graphite) + O2 (g) ----->CO2 (g) DeltaH= -394kl

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Answer 1

Answer #1

The information available is:

(1) Sr(s)+ 12O2(g) ---->SrO(s) DeltaH= -592kj

(2) SrO(s) + CO2(g) -----> SrCO3(s) DeltaH= -234kj

(3) C(graphite) + O2 (g) ----->CO2 (g) DeltaH= -394k

The change in enthalpy of reaction can be obtained from Hess Law.

So we need to rearrange the above given equations so that we can obtain the desired equation No. 4.

(4) Sr(s) + C(graphite) + 32 O2(g) -----> SrCO3(s)

The equation (4) can be obtanied by adding the above three equations

Sr + 1/2O2 + SrO + CO2 + C + O2 ---> SrO + SrCO3 + CO2

cancel the common terms on the two side of equation

Sr+ 3/2 O2 + C --> SrCO3 ( Our desired equation )

So Enthapy of reaction = DeltaH1 + DeltaH2 + DeltaH3 = -592kj + (-234kj) + (-394kJ) = -1220 KJ / moles

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Answer 2

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