Also, where will the equilibrium shift when NaHSO4 is added to it.\n\nFinally, is this reaction endothermic or exothermic?\n\n\nThanks!!! 10 points goes to best answer.
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I'm assuming you mean SO4-2, not O4-2. Anyway, Na2SO4 dissociates in water, increasing the concentration of SO4-2. This will shift the reaction to the left, to account for the increase in SO4-2.
426364
Let's take the HCl part first. Since HCl is an acid a solution of HCl contains lots of H+ ions. So when you add HCl the concentration of H+ ions increases making it more probable for H+ ions to colide and react with In- ions than before. To be more precise the reaction In- + H+ -> HIn will become more probable than before, so it will dominate over the HIn -> In- + H+ reaction until a new equilibrium is reached (equilibrium is characterized by the left to right and right to left reactions being equally probable). So in equilibrium the concentration of HIn will be higher after you have addeed HCl.\n\nIt's a very general equilibrium principle called leChateliers principle which comes into play here. The principle can be casted as follows: When you disturb an equilibrium (for instance by adding or removing one of the constituting components) the equilibrium will be shifted in a direction which tries to cancel the change. In this example: We add H+ and the equilibrium tries to remove some of the H+ ions again by shifting the equilibrium to the left (meaning that more H+ reacts with In- to get HIn).\n\nIf you add OH- instead the OH- will neutralize the H+ in the solution (acid + base gives water):\nH+ + OH- -> H2O\nso you are effectively removing H+ and according to leChateliers principle the equilibrium will try to create new H+ ions by shifting to the right (so you will have less HIn and more In- than before you added OH-). The microscopic explanation is again that removing H+ makes the reaction In- + H+ -> HIn less probable than before meaning that the reaction HIn -> In- + H+ will be dominating until a new equilibrium is reached so some HIn will effectively be turned into In-.\nHope this helps.
122250
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
515830
Part A:  equilibruim to the right because more co3 added\n\nPart B:  Equilibrum to left becausemore co2 added\n\nPart C:  Equilibrium to right to replace removed h20\n\nEquilibruim dynamic because although appears stable ions constantly moving back and forth just at the same time
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The reaction is as follows:\n\nCaCO3 + 2HCl -----------> CaCl2 + H2O + CO2\n\nThe reaction produces H2O & CO2 as biproducts and CO2 at room temperature will be a gas. So it will escape from the recation media (Contaner)\nAs a result, this reaction is not a reversible reaction (since one byproduct is missing) and hence as time goes water will adds up and HCl will dilute more and more ... rection kinetics slowed down.\n\nCaCl2 is very much  soluble in water.. so its not the reason\n\nIs this answer is enough to deserve 10 pts from you?\n\nHave fun in chemistry always!
185641
well, i do not know if this is correct but I'll give it a bash.\n\nfor your sulphuric acid there it undergoes a single H dissociation to create the HSO4- conjugate base (well i suppose technically it's amphoteric but for my purposes that doesn't matter) this is an extremely stable ion since it is resonance stabilised.  Thus it resists accepting back the proton and it resists losing it's other proton, because in doing so it would lose a degree of it's resonance.\n\nIn the group 2 hydroxides mentioned they obviously have ionic bonds and not covalent therefore they cannot have have resonance structures.  The reason they lose both hydroxides or none would be if the enthalpy of solution, that is if the polarity of the water molecule and the other factors of electron affinity/ionisation is strong enough to break the M--OH bond then it will do so twice and there is no way to stabilise a single dissocation product.\n\ni think.
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I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
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It is because you have the wrong compound (HSO3)!\n\n2 HI + K2SO3 ---> 2KI + H2SO3
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KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
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Na2S (ag) + 2 AgNO3 (ag) --> 2 NaNO3 (ag) + Ag2S (cr)\n\nB. Double displacement.\n\nA. Ba(NO 3)2 and H2O\n\nB. If there is a double arrow (pointing in both directions), then it is an equilibrium reaction.\n\nC. Ba(OH)2 and HNO3\n\nD. Barium\n\nE. 6 oxygen atoms. \n\nF. Ba(OH)2
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The second one:\nAF + H --> AFM + TM\n\nActivation E: +10KJ/mol is moderate\nHeat of Rx= +5.2 KJ/mol is positive, making this an endothermic reaction (requires the input of heat or energy to accomplish).\n\nSee the site below:
647359
Look up stuff like Arrhenius equations. I don't have my phys chem text book to hand.\n\nFor part two, I think it's D, as I'm fairly certain it's not any of the others.
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Na2S + 2 HCl > H2S + 2 NaCl  \n\nH2SO4 + BaOH > H20 + BaSO4  \n\nNH4Cl + NaOH > NaCl + NH4OH\n\nAgNO3 + NaCl > AgCl + NaNO3\n\nnot sure about complete or reversible..:)
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I think you are in my class. -0-... do you have chem test tommorow?\n\nI believe it is d.\n\nEquilibrium has to be in a closed system.\nTemperature changes K value so it is necessary to keep in a constant temperature.\nEquilibrium basically is defined as equal forward and reverse reaction reates.\n\nConcentration does not necessary has to be same.
365352
Correct.\nD. when delta G is positive the reaction is non-spontaneous
66446
the correct answer is B. A is wrong b/c it isn't an oxidation reduction reaction. Looking at a table of solubility products will confirm the answer is B.
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Basically what Le Chatelier's Principle is trying to say is that if a system in equilibrium experiences a change, then it will try and counteract that change.\n\nIn this case, the excess heat is the change experienced. To counteract the excess heat, some of it must be absorbed. Thus, the endothermic reaction in which heat is absorbed will occur at a higher rate than the other reaction. \n\nThe endothermic reaction is the forward reaction. (Equilibrium shifts to the right.) Thus, more N2O4 will be converted to NO2 than the other way round, resulting in less brown-coloured N2O4 molecules in the system. Therefore, the intensity of the colour will decrease.\n\n:) hope you find this useful!