Calculate the Gibbs Free Energy of the reaction shown below: 12 NH3(g)   +   21 O2(g)    →  8 HNO3(g)   +   4 NO(g)   +   14 H2O(g)   USE only the SIGN & NUMBER for the answer but include FOUR DIGITS.

(b)  Calculate the enthalpy change for the reaction C(graphite) + 2 H2(g) → CH4(g) with the help of the enthalpy changes of the following reactions: C(graphite) + O2(g) → CO2(g) ∆H= -393.5 kJmol−1 H2(g) +1/2O2(g) → H2O(l) ∆H = -285.8 kJmol−1 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H = -890.4 kJmol−1 The temperature is 25 ◦C in all instances.

In Part B of the experiment, a student mixes 30.0 mL of 1.1 M HC1(aq) with 30.0 mL of 1.000 M NaOH(aq) in a well-insultated calorimeter and observes that the temperature of the solution increases by 5.75 °C. These are similar conditions as the previous two problems. How many moles of water are formed? number of moles of water = mol

Given the following Fe,03 (s) + 3 CO (g) →2 Fe (s) + 3 CO2 (g) AG' = - 29.4 kJ 3 Fe203 (s) + Co (g) →2 Fez04 (s) + CO, (g) AG° = -61.6 kJ Calculate AG° for Fez04 (s) + CO (g) → Fe (s) + Fe,03 (s) + CO2 (g)

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

How much heat must be removed from a 90.5 g sample of benzene (M = 78.11 g mol-1) to cool the sample from 305 K to 215 K?  The following physical data may be useful.ΔfusH = 9.8 kJ mol-1ΔvapH = 30.8 kJ mol-1Cs(l) = 1.73 J g-1 K-1Cs(s) = 1.51 J g-1 K-1Cs(g) = 82.4 J g-1 K-1 Tm = 279.0 KTb = 353.3 K Note: The word 'removed' in the question already accounts for the negative sign involved in the heat transfer.   Express your answer to three significant figures.

Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. 2CH3OH(g)+3O2(g)----> 2CO2(g)+4H2O(g)

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)

Calculate ΔH° for the reaction shown below using standard enthalpy of formation values. KCl (s)   →   K+1 (aq)   +   Cl-1 (aq) Be sure to include a unit with your answer.