175Q16
Because the number of moles of gas are increasing from 6 to 7 in the reaction shown below, at constant pressure ΔE is predicted to be slightly ________ negative than ΔH.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
Δn = total no of moles of products - total no of moles of reactants in gases
= (3 + 4) - (1 + 5) = 1
ΔH = ΔE + ΔnRT
ΔH = ΔE + 1RT
ΔE = ΔH -RT
at constant pressure ΔE is predicted to be slightly RT negative than ΔH.
175Q16 Because the number of moles of gas are increasing from 6 to 7 in the...
Calculate the volume change (in Liters) during the combustion of propane in which ΔE = -3083.67 kJ and ΔH = -3087.12 kJ at a constant pressure of exactly one atm and constant temperature. C3H8(g) + 5 O2(g) ↔ 3 CO2(g) + 4 H2O(l) The product of pressure and volume change should give units of L*atm. It can be shown that 1 L*atm = 101 J.
Entropy is likely to decrease for the reaction below, because there are 7 moles of gas on the reactant side and no moles of gas on the product side. 4 C(s) + 6 H2(g) + O2(g) → 2 C2H5OH(l) TRUE or FALSE
When 20.00 moles of H2(g) reacts with 10.00 mol of O2(g) to form 20.00 mol of H2O(l) at 25°C and a constant pressure of 1.00 atm. If 1366 kJ of heat are released during this reaction, and PΔV is equal to -74.00 kJ, then ΔH° = -1366 kJ and ΔE° = -1292 kJ.ΔH° = +1366 kJ and ΔE° = +1440 kJ.ΔH° = -1366 kJ and ΔE° = -1440 kJ.ΔH° = +1366 kJ and ΔE° = +1292 kJ.
At constant pressure for which of the reactions shown below should ΔH° be greater than ΔE° ? I. 2 SO2(g) + O2(g) → 2 SO3(g) II. C23H48(g) + 35 O2(g) → 23 CO2(g) + 24 H2O(l) III. H2(g) + Cl2(g) → 2 HCl(g) IV. N2O4(g) → 2 NO2(g)
When the chemical reaction shown below is at equilibrium, if the volume is reduced, in what direction would the reaction shift? C3H8(g) + 5 O2(g) ⇌ 3 CO2(g) + 4 H2O(l) ΔH° = −2220 kJ a. To the left b. To the right c. No change Would it shift to the left? Or right since volume works inversely of pressure
When .514 g of C12H10 undergoes combustion in a bomb calorimeter, the temperature increases from 25.8oC to 29.4oC. The heat capacity of the bomb and all of its contents is 5.86 kJ/oC. What is the heat released per mole of C12H10? Is the number you calculated ΔE or ΔH? Why? Is ΔE = ΔH for this reaction? Why? Balanced reaction: 2 C12H10 (s) + 29 O2 (g) → 24 CO2 (g) + 10 H2O (g) If ΔE and ΔH are...
The enthalpy change, ΔH, for a reaction at constant pressure is defined as: ΔH = ΔE + PΔV. For which of the following reactions will ΔH be approximately equal to ΔE? Select all that apply. Group of answer choices 2 NO2(g) -> N2(g) + 2 O2(g) Ca(OH)2(aq) + H2SO4(aq) -> 2 H2O(l) + CaSO4(s) C(s) + O2(g) -> CO2(g) None of the above
Consider the following reaction at equilibrium. What effect will increasing the temperature have on the system? C3H8(g) + 5 O2(g) ⇌ 3 CO2(g) + 4 H2O(l) ΔH° = −2220 kJ. a. The reaction will shift to the right and K increases. b. No effect will be observed. c. The reaction will shift to the right and K decreases. d. The reaction will shift to the left and K increases. e. The reaction will shift to the left and K decreases.
The combustion of propane gas (C3H8) is used to fuel barbeque grills. In order to produce 5.65 moles of water how many moles of oxygen gas are needed? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l
un plase uansion from a liquid to a gas • an increase in the number of moles of a gas during a chemical reaction EXAMPLE 17.1 Predicting the Sign of Entropy Change Predict the sign of AS for each process: (a) H2O(g) → H2O(1) (b) Solid carbon dioxide sublimes. (c) 2 N2O(g) —> 2 N2(8) + O2(g) SOLUTION (a) Since a gas has a greater entropy than a liquid, the entropy decreases and AS is negative. (b) Since a solid...