Under what conditions will the following reaction be the most product-favored?
N2O(g) + NO2(g) <==> 3 NO(g) ; H = + 156 kJ
A. low pressure, low temperature B. high pressure, low temperature C. high pressure, high temperature D. low pressure, high temperature
D.) The above reaction given is an endothermic reaction because here Delta H given is positive . Endothermic reaction are those reactions which absorbs heat.
This reaction will be most product favoured at high temperature and low pressure .
Effect of high temperature: An endothermic reaction absorbs heat from the sorroundings for the formation of products. When temperature is increased ,it increases the amount of product formed ,and decreases the amount of reactants. So high temperature favours the product based endothermic reaction whereas reactant based endothermic is favoured by low temperature.
Effect of pressure: Effect of pressure on a reaction is due to number of moles of gas. Because in case of gases, with decrease in pressure, volume increases . Here number of moles of gas are more on product side than the reactant side. So the product based reaction will be favoured by low pressure . This is explained by le chatelier principal.
Under what conditions will the following reaction be the most product-favored? N2O(g) + NO2(g) <==> 3...
Consider the following reaction: C 2H 2( g) + 2H 2( g) C 2H 6( g) Δ H° rxn = –311 kJ Under which conditions would the product yield be at a maximum? A. high temperature, high pressure B. low temperature, high pressure C. high temperature, low pressure D. low temperature, low pressure \E. none of the above, unless a catalyst is present
3. Calculate AH for the reaction Ato AL120103 N2O(g) + NO2(g) → 3 NO(g) from the following enthalpies of reaction: NO(g) + O2(g) + NO2() N2O(g) + N2O) + O2(8) 1 N2(g) + O2(g) → NO(g) AH = -56.6 kJ AH = -81.6 kJ AH = +90.4 k]
Given the following equation, N2O(g) + NO2(g) → 3 NO(g) AG°/x = -23.0 kJ Calculate AGºrxn for the following reaction. CHO COURO 3 NO(g) + N2O(g) + NO2(g) 153 ou te ato O bbou n esto bancabies A) -23.0 kJSOU nude au CODE B) 69.0 kJ e je bom su se 159 Hou e ab C) -69.0 kJ D) -7.67 kJ ost adapun E) 23.0 kj
For the reaction 2 NO(g) + O2(g) 2 NO2(g) G° = -71.1 kJ and S° = -146.5 J/K at 294 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 294 K. The standard enthalpy change for the reaction of 1.67 moles of NO(g) at this temperature would be kJ.
For the reaction: N2O (g) + NO2 (g) <--> 3 NO(g) , Kc = 4.2 x 10-4 at 500. K. What is Kp for the reaction at this temperature?
Find the value of the equilibrium constant, Kp, for the following reaction N2O(g) + NO2(g) → 3NO(g) at 900 K given the following data: N2O(g): AGfº=149.2 kJ/mol; NO2(g): AGfº=89.4 kJ/mol; NO(g): AGfº=79.0 kJ/mol. 0.192 1.21 1.00 0.826 O 1.43
Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ Calculate ΔG°rxn for the following reaction. 18 NO(g) → 6 N2O(g) + 6 NO2(g) -3.83 kJ -23.0 kJ 138 kJ -138 kJ 23.0 kJ
Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ Calculate ΔG°rxn for the following reaction. 18 NO(g) → 6 N2O(g) + 6 NO2(g) -3.83 kJ -138 kJ 23.0 kJ -23.0 kJ 138 kJ
i need help with these questions
1. Under the following reaction conditions, what is the favored form of compound shown? NAOMe MeOH A. The Enol is Favored B. The Enol and Ketone are present (50:50 mixture) D. The Enolate is Favored C. The Ketone is Favored 2 What reagent best finishes the following reaction scheme. A. Br. HBR (cat) B. NaOH, Br C. HB, ROOR D. NBS, hv 3. What is the correct product for the following reaction? H2SO4 3-methyl-1-butanol...
8) The rate law for the following reaction N2O(g) + NO2(g) → 3NO(g) is Rate = k [N2O][NO2]2. Fill in the blank in the table below. Experiment [N2O], M [NO2], M Initial Rate, MS-1 1 0.100 0.100 2.0x10-3 2 0.300 0.100 6.0x10-3 3 0.100 0.200 ? a) 2.0x10−3 Ms−1 b) 4.0x10−3 Ms−1 c) 8.0x10−3 Ms−1 d) 6.0x10-3 Ms-1 e) 1.0x10−2 Ms−1