This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's how to calculate the heat of combustion at constant pressure: Step 1: Write the balanced chemical equation for the combustion of ethene (alkylene, C_2H_4). Combustion of ethene with oxygen produces carbon dioxide and water. Since water is specified to be in the liquid state, the equation is: C_2H_4(g) + 3O_2(g) 2CO_2(g) + 2H_2O(l) Step 2: Calculate the change in the number of moles of gaseous substances ( n_g). n_g is the moles of gaseous products minus the moles of gaseous reactants. n_g = (moles of gaseous products) - (moles of gaseous reactants) From the balanced equation: Gaseous products: 2 moles of CO_2(g) Gaseous reactants: 1 mole of C_2H_4(g) + 3 moles of O_2(g) = 4 moles n_g = 2 - (1 + 3) = 2 - 4 = -2 mol Step 3: Convert the given temperature to Kelvin and the gas constant R to kJ K^-1 mol^-1. Given temperature T = 17^. T(K) = 17 + 273.15 = 290.15 K Given gas constant R = 8.314 J K^-1 mol^-1. R = 8.314 J K^-1 mol^-1 × 1 kJ1000 J = 0.008314 kJ K^-1 mol^-1 Step 4: Calculate the heat of combustion at constant pressure ( H) using the relationship between H and U. The heat of combustion at constant volume ( U) is given as -1389.55 kJ. The formula relating H and U is: H = U + n_g RT Substitute the values: H = -1389.55 kJ + (-2 mol) × (0.008314 kJ K^-1 mol^-1) × (290.15 K) H = -1389.55 kJ - (2 × 0.008314 × 290.15) kJ H = -1389.55 kJ - 4.8248 kJ H = -1394.3748 kJ Rounding to two decimal places: H = -1394.37 kJ That's 2 down. 3 left today — send the next one.