Question

Calculate δhrxn for the following reaction: sro(s) co2(g)→srco3(s) use the following reactions and the given values of δh for them: sr(s) co2(g) 1/2 o2(g)→srco3(s), δh =−819.5kj, 2sr(s) o2(g)→2sro(s), δh=−1096.0kj

Answer 1

Using Hess's Law, the enthalpy change for the reaction SrO(s) + CO2(g) -> SrCO3(s) was calculated by adjusting and summing up given reactions, resulting in ΔHrxn being -271.5 kJ.

Step-by-step explanation:

To calculate the enthalpy change (ΔHrxn) for the reaction SrO(s) + CO2(g) → SrCO3(s), we need to the use Hess's Law which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps into which the reaction can be divided. We have been given the following reactions:

• Sr(s) + CO2(g) + ½ O2(g) → SrCO3(s), ΔH = -819.5 kJ
• 2 Sr(s) + O2(g) → 2 SrO(s), ΔH = -1096.0 kJ

To find the enthalpy change for the formation of SrCO3 from SrO and CO2, we can start by reversing the second reaction to represent the decomposition of SrO, which also changes the sign of ΔH:

• 2 SrO(s) → 2 Sr(s) + O2(g), ΔH = +1096.0 kJ

Dividing this reaction by 2 to suit the coefficients of SrO(s) in our desired reaction:

• SrO(s) → Sr(s) + ½ O2(g), ΔH = +548.0 kJ

The first given reaction is correct as is. Now, if we add the two adjusted reactions, the Sr(s) and ½ O2(g) cancel out and we are left with:

SrO(s) + CO2(g) → SrCO3(s)

The sum of the enthalpy changes for our reaction (ΔHrxn) is the sum of the enthalpy changes of the steps, which is:

ΔHrxn = (+548.0 kJ) + (-819.5 kJ) = -271.5 kJ

Therefore, the ΔHrxn for the reaction SrO(s) + CO2(g) → SrCO3(s) is -271.5 kJ.