Question

Standard heats of formation for reactants and products in the reaction below are provided. 2 HA(aq) + MX2(aq) → MA2(aq) + 2 HX(l) Substance ΔHf° (kJ/mol) HA(aq) 280.623 HX(l) 100.27 MA2(aq) 131.46 MX2(aq) -131.718 What is the standard enthalpy of reaction, in kJ? Report your answer to three digits after the decimal.

Answer 1

Solution:-

​​​​​​We know that,

∆H^{​​​​​​}° = ∆H_f° (products) - ∆H_f° (reactants)

Where,

∆H° = standard enthalpy of reaction

∆H_f° (products) = Total standard heat of formation of products

∆H_f° (reactants) = Total standard heat of formation of reactants.

Now,

∆H° = [2 × ∆H_f° (HX) + 1 × ∆H_f° (MA₂)] - [2 × ∆H_f° (HA) + 1 × ∆H_f° (MX​​​​​​​​​​​​​₂)]

∆H° = [2 × 100.27 + 1 × 131.46] - [2 × 280.623 + 1 × (-131.718)]

∆H° = [200.54 + 131.46] - [561.246 - 131.718]

∆H° = 332 - 429.528

∆H° = -97.528 kJ

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