Answer completely Calculate the standard enthalpies of combustion at 25ᵒC of the following: (a) C4H8(g) + H2(g) à C4H10(g) ΔHᵒ = -2878.6kJ (b) C4H6(g) + 2H2(g) à C4H10(g) ΔHᵒ = -2543.5 kJ
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- Select ALL reactions below that the enthalpy change under standardconditions correspond to a standard enthalpy of formation? 1/2 H2(g) + 1/2 N2(g) + 3/2 O2(g) --> HNO3(g) 2H2(g) + C(s) --> CH4(g) CO2(g) + C(s) --> 2CO(g) SO2(g) + 1/2 O2(g) --> SO3(g) N2(g) + 2O2(g) --> 2NO2(g) Mg(s) + 1/2 O2(g) --> MgO(s) CO2(g) + H2(g) --> CO(g) + H2O(g) CO(g) + H2O(g) --> CO2(g) + H2(g)Please help me with my hw. Tysm❤️ ACTIVITY: Answer completely. Calculate the standard enthalpies of combustion at 25ᵒC of the following: (a) C4H8(g) + H2(g) à C4H10(g)ΔHᵒ = -2878.6kJ (b)C4H6(g) + 2H2(g) à C4H10(g)ΔHᵒ = -2543.5 kJΔH = +572 kJ for the decomposition of water by the reaction 2 H2O() → 2 H2(g) + O2(g). How many grams of water can be decomposed by 5,932 kJ of energy? (Hint: standard unit for heat from a reaction is KJ/mol and FW for water is 18.01g/mol) (3 SF)
- Calculate the heat of reaction, Hrxn, for the combustion of one mole of C6H6(l) using the ∆reactions provided below. 6 C(s, graphite) + 3 H2(g) → C6H6(l) ∆H = 49.0 kJ H2(g) + ½ O2(g) → H2O(l) ∆H = -285.8 kJ∆ C(s, graphite) + O2(g) → CO2(g) ∆H = -393.5 kJ I've attached a photo for more clarity. Thanks!Given the following enthalpies of combustion: (a) 2H2(g) + O2(g) --> 2H2O(g); DH = -483.6 kJ (b) 2C3H6(g) + 9O2(g) --> 6CO2(g) + 6H2O(g); DH = -3853.6 kJ (c) C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l); DH = -2043.7 kJ Calculate DH for the following reaction: (A 4) C3H6(g) + H2(g) --> C3H8(g) (A) -124.9 kJ (B) -366.7 kJ (C) -2293.5 kJ (D) -4262.3kJIf the needed reaction is 4C --> D and the given reaction is D --> 4 C with ΔH = 134.7 kJ, what is the ΔH for the needed reaction? Answer in kJ without the units.
- Given the following information: Chemical Reaction: 2 C10H10 (l) + 25 O2 (g) → 20 CO2 (g) + 10 H2O(l) ∆H = 10,314 KJ/mol The standard heat of formation (KJ/mol) at 298.15 oC : CO2 = -393.5 ; H2O(l) = -285.8 What is the standard enthalpy of formation of C10H10?Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) → 2CO2(g) + 2 H2O(l) ΔH° = -1411 kJ ;C(s) + O2(g) → CO2(g) ΔH° = -393.5 kJ ;H2(g) + 1/2 O2(g) → H2O(l) ΔH° = -285.8 kJ Question 2 options:Automobiles and trucks pollute the air with NO. At 2000.0°C, Kc for the reaction is 7.42 × 10–4, and ΔH for the reaction is 180.6 kJ.N2(g)+O2(g)2NO(g) What is the value of Kc at 1000.0°C?
- Please don't handwrite solution. 5) What quantity of heat (in kJ) will be absorbed by a 31.5 g piece of aluminum (specific heat = 0.930 J/g･ °C) as it changes temperature from 23.0 °C to 67.0 °C? 7) When 1.831 grams of sucrose (Molar mass 342.3 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter increases from 22.410 °C to 26.630 °C. If the heat capacity of the calorimeter is 4.900 kJ/ °C, what is the heat of combustion of sucrose, in kJ/mol?When gasoline burns in an automobile engine, the heat released (316 J) causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the work associated with the expanding gases is equal to 453 J, how much is the change in internal energy of the gases? If answer is negative, include sign, e. g. -1000. No need to add positive sign if answer is positive, e.g. 1000.a) Benzoic acid, C7H6O2, occurs naturally in many berries. Suppose you burn 1.500 g ofthe compound in a combustion calorimeter and find that the temperature of thecalorimeter increases from 22.50 °C to 31.69 °C. The calorimeter contains 775 g ofwater, and the bomb has a heat capacity of 893 J/K. How much heat is evolved permole of benzoic acid? b) Refer to problem 11 above. The balanced equation for the combustion of benzoicacid is 2 C7H6O2 (s) + 15 O2 (g) → 14 CO2 (g) + 6 H2O (l). Calculate ∆E and ∆H for the above reaction as written. Assume the reaction occursat 31.69°C. ---------------------------------------------------------------------------------------------------------------------- Answer: a) - 3.09 × 10^3 kJ/mol b) ∆E = - 6.18 × 10^3 kJ w = + 2.534 kJ ∆H = - 6.18 × 10^3 kJ