Consider the following chemical reactions and their standard reaction enthalpies (ΔH) at 298 K. NH + HCl(g) → NHCl(s) ΔH = -176 kJ/mol N(g) + 3 H(g) → 2 NH(g) ΔH = -92 kJ/mol N(g) + 4 H(g) + Cl(g) → 2 NHCl(s) ΔH = -629 kJ/mol   Use this thermochemical data to calculate the standard heat of reaction for the synthesis of hydrogen chloride gas from H(g) and Cl(g) at a temperature of 298 K. The value of ΔH (at 298 K) for the reaction H(g) + Cl(g) → 2 HCl(g) is:

Calculate the ΔHo for the reaction, 6NO2(g) + 8NH3(g) → 7N2(g) + 12H2O(l).   N2(g) + 2O2(g) → 2NO2(g)   ΔHo=+66.36 kJ 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)   ΔHo=-1530.36 kJ

Calculate ΔHrxn for the reaction: 7 C (s) + 8 H2 (g) → C7H16 (g) The following reactions and associated ΔHrxn values are may likely be helpful to you: Reaction A: C7H16(g) + 11 O2 (g) → 7 CO2 (g) + 8 H2O (g) ΔHrxn = -4817 kJ Reaction B: C (s) + O2 (g) → CO2 (g) ΔHrxn = -393.5 kJ Reaction C: 2 H2 (g) + O2 (g) → 2 H2O (g) ΔHrxn = -483.6 kJ

55. Does the standard enthalpy of formation of H2O(g) differ from ΔH° for the reaction 2H2(g)+O2(g)⟶2H2O(g)?

Given the information below, what is ΔGrxn for the reaction CO(g) → C(s) + ½ O2(g)?C(s) + O2(g) → CO2(g)     ΔGrxn = –394.4 kJCO(g) + ½ O2(g) → CO2(g)     ΔGrxn = –257.2 kJ

If we measure the heat change (q) of everything in a system except the calorimeter, we can use the first law of thermodynamics (∆Euniverse = 0) to calculate the heat change of the calorimeter (qcal).  In this reaction, we combine 5.85 g of 63.7 ºC water with 6.63 g of 18.9 ºC water in a calorimeter.  Since some of the heat is absorbed by the calorimeter, the final temperature of the water is 28.8ºC.  Since the hot and cold water are combined, the final temp of the water is the final temp for the hot water and the cold water.  Calculate the heat change of the calorimeter in J. Remember that all of the heat changes must add up to zero so you can calculate the heat change of the hot water (qhotwater) and the heat change of the cold water (qcoldwater) using for each where Cs = 4.186 J/gºC.  The first law of thermodynamics tells us that all of these changes sum up to zero so   Enter your answer to one decimal place (tenths)

Calculate the change in internal energy (ΔE) for a system that is giving off 3.0 kcal of heat and is changing from 1.00 L to 8.00 L in volume at a pressure of 2.00 atm. (Remember that 101.3 J = 1 L∙atm; 1 cal = 4.18 J)

Given the following reaction: 2NO (g) + O2 (g) → 2NO2 (g)           ΔrH0 = - 114.1 kJ/mol Calculate the enthalpy change when 1.25 g of NO gas is converted completely converted to NO2 gas. Enter your answer to 2 decimal places.

The ΔHo reaction for the oxidation of ammonia4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) is -905.2 kJ.Calculate ΔH̊ for ammonia. The standard enthalpies of formation are:NO(g) = +90.3 kJ/molH₂O(g) = -241.8 kJ/mol.