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What is the standard enthalpy change for the reaction between 50 mL of 1 M sulfuric acid and 40 mL of 1.5 M sodium hydroxide at 25 °C, given that the temperature change observed was 6.5 °C?

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To calculate the standard enthalpy change for the reaction, we need to first determine the heat released or absorbed during the reaction  q  and then divide it by the moles of the limiting reactant.The balanced chemical equation for the reaction between sulfuric acid  H2SO4  and sodium hydroxide  NaOH  is:H2SO4 + 2 NaOH  Na2SO4 + 2 H2OFirst, we need to determine the limiting reactant. We can do this by calculating the moles of each reactant:moles of H2SO4 = volume  concentration = 50 mL  1 M = 50 mmolmoles of NaOH = volume  concentration = 40 mL  1.5 M = 60 mmolSince 1 mole of H2SO4 reacts with 2 moles of NaOH, we can see that H2SO4 is the limiting reactant because:50 mmol H2SO4   2 moles NaOH / 1 mole H2SO4  = 100 mmol NaOH requiredNow, we can calculate the heat released or absorbed during the reaction  q  using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity of the solution  assuming it's similar to water, which is 4.18 J/gC , and T is the temperature change.The total volume of the solution is 50 mL + 40 mL = 90 mL, and assuming a density of 1 g/mL  similar to water , the mass of the solution is 90 g.q =  90 g    4.18 J/gC    6.5 C  = 2449.7 JNow, we can calculate the standard enthalpy change  H  by dividing the heat released or absorbed  q  by the moles of the limiting reactant  H2SO4 :H = q / moles of limiting reactant = 2449.7 J / 50 mmol = 49.0 J/mmolSo, the standard enthalpy change for the reaction between 50 mL of 1 M sulfuric acid and 40 mL of 1.5 M sodium hydroxide at 25 C, given that the temperature change observed was 6.5 C, is approximately 49.0 J/mmol.

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