Data: Data: g of HCl g of Mg/MgO Temp HCl Highest Temp. Reaction 1 100.09gHCl 1.00g MgO 24.0 C 28.8 C Reaction 2 99.55gHCl 0.50g Mg 23.7 C 46.0 C Calculations: q or ∆ H = mc ∆ T **C is a constant = 4.18j/g c Reaction 1: MgO: m = 1.00g MgO + 100.09g HCl = 101.09g ∆ T = 28.8 C – 24.0 C = 4.8 C ∆ H = 101.09g * 4.18j/g c * 4.8 C = 2028j / 1000j = 2.028Kj 1.00gMgO *1 molMgO/40.31gMgO = 0.025mol 2.028Kj/0.025mol = 81.12Kj/mol ---but since it is being released it is negative : -81.12Kj/mol
Reaction 2: Mg m = 0.50gMg + 99.55gHCl = 100.05g ∆ T = 46.0 C – 23.7 = 22.3 C ∆ H = 100.05g * 4.18j/g c * 22.3 C = 9326.5j / 1000j = 9.326kj 0.50gMg* 1mol Mg/ 24.39gMg = 0.021mol Mg 9.326kj/0.021mol = -444.1kj/mol Balanced equations: 2HCl + MgO H 2 O +MgCl 2 2HCl + Mg H 2 + MgCl 2 H 2 + 1 2 O 2 H 2 O ∆ H f for H 2 O was found in the book, = -285.8Kj/mol Analysis: Mg + 1 2 O 2 MgO ∆ H = -648.78kj/mol 2HCl + MgO H 2 O +MgCl 2 ∆ H = -81.12Kj/mol H 2 O +MgCl 2 2HCl + MgO ∆ H = 81.12Kj/mol 2HCl + Mg H 2 + MgCl 2 ∆ H = -444.1kj/mol H 2 + 1 2 O 2 H 2 O ∆ H = -285.8Kj/mol ∆ H = 81.12kj/mol + (-444.1kj/mol) + (-285.8kj/mol) = ∆ H of Mg + 1 2 O 2 MgO = -648.78kj/mol Conclusion: ∆ H f for MgO is =-648.78kj/mol, which was obtained through Hess’ law. The actual value of ∆ H f for MgO is -601.6kj/mol. The percent error = -648.78kj/mol - -601.6kj/mol / -601.6kj/mol * 100 = 7.84% The percent error within the experiment was only 7.84%. Some potential sources of error could be the heat of reaction one affecting the heat of reaction two due to leftover heat tempering with the starting heat of reaction two. Another odd occurrence that may have affected the outcome of this experiment was the calorimeter weight seemed to have changed between reaction one and two despite it being the same device used in both reactions. The value obtained was close to the actual value and was found using Hess’ law and actually completing the reactions to find data in order to obtain the equations to perform Hess’ law to find it.
- Fall '19
- Mrs. Collias