When 1.50 mol of CH4(g) reacts with excess Çl2(g) at constant pressure according to the chemical equation shown below, 1062 kJ of heat are released. Calculate the value of AH for this reaction, as written. 2 CH4(g) + 3 Cl2(g) → 2 CHC13(1) + 3 H2(g) ΔΗ- A) -1420 kJ B) -708 kJ C) +708 kJ D) +1420 kJ Answer: A 1.50mal CHy 10GR KJ,|| 3mol Clz 15mol CHy -796.5KJ Rmol CHy

When 1.50 mol of CH4(g) reacts with excess Çl2(g) at constant pressure according to the chemical equation shown below, 1062 kJ of heat are released. Calculate the value of AH for this reaction, as written. 2 CH4(g) + 3 Cl2(g) → 2 CHC13(1) + 3 H2(g) ΔΗ- A) -1420 kJ B) -708 kJ C) +708 kJ D) +1420 kJ Answer: A 1.50mal CHy 10GR KJ,|| 3mol Clz 15mol CHy -796.5KJ Rmol CHy

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

determine the pOH of a 0.0453 mol L-1 KOH solution at 25 ℃
Calculate the value of Δ?∘rxnΔGrxn∘ for the following reaction at 278 K. Ka = 2.9 × 10–8 and assume Ka does not change significantly with temperature. HClO(aq) + H2O(l) <--> ClO-(aq) + H3O+(aq)
Ka = 4.5 × 10-4 at 25°C of nitrous acid (HNO2). What is the ΔG value at equilibrium?
Please calculate the Ka for a weak acid given the following data: pH = 3.72, and an initial acid concentration of 1.43 x 10-4 M (note: please express your answer with 3 sig figs.; the equilibrium reaction is shown below:) НА(аq) + H,0 () H;O* (aq) + A´ (:
The following reaction has a AG° value of +8.43 kJ/mol at 25 °C. HA(aq) + H20(1) =H;O*(aq) + A (aq) Calculate the KA for the acid HA.
Hydrogen peroxide, H₂O₂, has a density of 1.45 g/cm³ and decomposes spontaneously but slowly. The decomposition reaction will proceed at a faster rate if iodide ion, a catalyst, is added. The steps for the catalyzed reaction are: H₂O₂(aq) + (aq) → H₂O(l) + Ol-(aq) OO H₂O₂(aq) + Ol(aq) → H₂O(l) + O₂(g) +1-(aq) a. What is the net reaction? b. If 10.00 mL of H₂O₂ undergoes this reaction, what mass of oxygen is produced? c. If the reaction has an 85.5% yield, what mass of oxygen is actually produced?
Acetaldehyde (CH3CHO) is an important chemical both industrially and biologically. For instance, it is a (somewhat toxic) intermediate in the body's metabolism of ethanol into acetic acid, and thus is possibly implicated in the "hungover" symptoms of someone who has had too much to drink the night before. In aqueous solution, it establishes an equilibrium with a hydrated form, shown below. CH3CHO (aq) + H2O (l) <--> CH3CH(OH)2 (aq) You start with an aqueous sample, already at equilibrium, with the CH3CH(OH)2 (the hydrated form) at a concentration of 2.60 M. You have no information about how much, if any, of the anhydrous form (CH3CHO) is initially in the flask. If you add 2.0 M of CH3CHO to the reaction flask, and as the equilibrium is being restored the amount of CH3CH(OH)2 changes by 1.13 M, what is the final amount of CH3CHO?