what happens to the sign of specific heat if a reaction is run reverse direction from the way it is written?
341402
Specific heat does not apply to reactions, only actual chemicals.  You are probably thinking of the heat of reaction, or enthalpy.  The change in enthalpy of a reaction (written as delta H, delta as the Greek letter that looks like a little triangle) flips its sign when the reaction is run in the reverse direction.  So a positive delta H value becomes negative when the reaction is run in reverse direction, and a negative delta H value becomes positive.
460906
DH reaction = DH products – DH reactants\n\nExample:  2A + B -- C\nDH = DH C – (2 DH A + DH B)\n\nIf DH is negative: exothermic reaction \nIf DH is positive: endothermic reaction
689177
The equation you wrote above is an exothermic reaction, because it is going from unstable reactants to more stable products.  You can usually tell something is spontaneous and exothermic when you have water as a product.  This is due to the fact that water is one of the most stable compound on earth and that is the reason why most of earth's surface is comprised of water.   \n\nGiven this, it should be a spontaneous reaction going forward in STP.  However since it is an exothermic reaction, given high temperature of 1000K, the reaction will push back to the reactant side, thus going backwards according to Le Chatelier's Principle.  Showing Eqution:\n\n4HCL(g) + O2(g) <---> 2CL2(g) + 2H2O(g) + heat\n\nIf you add more Heat in this reaction, the reaction will move <--.
731401
Correct.\nD. when delta G is positive the reaction is non-spontaneous
328113
It's not a reaction, but rather the dissolution of a salt (usually ammonium nitrate) in water.  When you break the inner water pack, the ammonium nitrate dissolves.  This process for this particular salt is highly endothermic, that is it absorbs heat, reducing the temperature of the pack.\n\nA reverse reaction happens w/ instant heat packs, except us the salt precipitates (solidifies), an exothermic process, giving off heat.
897593
If you are doing this in terms of the system,\n\nsince the system is losing energy (delivers heat) then the amount of energy is "-1275 J".  The surrounding add heat (perform work on) to the system, so the amount of energy is "+855".  The internal energy of the system is the sum of the energy loss and the work done to the system, so -1275 J + 855 J = -420 J.\n\nIf you are doing this problem in terms of the surroundings, the all the signs are reversed and you should end up with +420 J.
178651
Your question is not clear but I take it you want 2 know what happens when heat is applied to a chemical reaction. \n\nHeat increases kinetic energy in molecules. If you apply heat, bombardment of molecules increase and therefore the reaction is favoured. In terms of equilibrium in reversible reactions, the forward reaction is favoured.
450334
Okay, cool!\nThe diagram is set up so that on the horizontal x-axis is the "reaction coordinate," which is basically just time as the reaction proceeds.  The vertical y-axis is energy.  \n\nThe potential energy of the reactants is where the graph curve starts, touching the y-axis.  The potential energy value is the y-value at this location.  \n\nWhere the graph curve levels off at the end is the products, so the potential energy of the products is the y-value where the curve levels off.  \n\nThe activation energy is how much energy had to be added into the system to get to the activated complex, located at the peak of the diagram's curve.  Take the y-value at the top of the curve and subtract from the the potential energy of the reactants.  \n\nThe heat of reaction is the difference in the potential energies of the products and reactants.  Take the potential energy of the products, and subtract from that the energy of the reactants.  \n\nIf a catalyst were added, the potential energy of the products and reactants remain unchanged, but the activation energy is less, so you draw a curve that starts in the same place and ends in the same place as the original diagram, but your curve just won't go up as high.  The hill for a catalyst is less steep.
403536
The sign of delta G can be figured out qulitatively.\n\nIf a certain reaction is spontaneous, then delta G is negative but if a certain reaction is non-spontaneous, then delta G will have a positive sign.
896452
Not sure the level with which you are learning, so feel free to ignore the thermodynamics explanation.  If you just want to answer the question, select D.\n\nLike most test questions, there is not enough conditions specified and truly, we could choose several anwers, but only after making assumptions that less likely to be in the testers mind.  I will assume a high school or intro level college course because above that this should be challenged as too vague.\n\nUse logic to realize that answer choice C is not true. Evaluating the others we realize that A means only heat is given off when bonds form and B means heat is absorbed or added during the formation of bonds.  D means conditions exist in which either situation is possible and E means that final product is in a state that has a higher value for entropy.  This only holds true when both Temp and pressure of a system is held constant, but we will eliminate later as well.\n\nBasically a bond can occur when the system goes from a high energy state to a lower energy state.  Formation of bonds is essentially a chemical reaction.  The Gibbs Free Energy equation:\n\n (delta)G=(delta)H-T(delta)S\n\nT=temp of system in Kelvin\n(delta)H is the heat energy change\n(delta)S is the change in entropy (disorder)\n\ndefines whether a reaction will occur spontaneously or not.  (deltaG) must be negative for this to be true. So we will assume \nusing this equation and recognizing that (delta)H tells us exo or endothermic if it is positive or negative , examination of the possibilities of (delta)H and (delta)S,\n\nIf (delta)H is negative and (delta)S is positive the equation will result in a negative value for (delta)G which means a favorible condition exists for bonding to occur.  Therefore, the condition of (delta)H negative (energy leaves as heat - exothermic) is true.  We can thus eliminate answer choice B.\n\nIf the opposite is true and (delta)H is positive and (delta)S is negative, the result is (delta)G that is positive and therefore not spontaneous.  no reaction or bonding occurs.\n\nIf both (delta)H is positive and (delta)S is positive, only a large enough (delta)S will overcome the positive (delta)H in the equation.  So if both are positive and T(delta)S is greater than (delta)H the bonding will occur.  Therefore, when (delta)H is positive (endothermic) the reaction may occur as well.  We can thus eliminate choice A.\n\nWe have shown that C is a true statement because conditions can be met in which the reaction is both exo or endothermic.  However, please note that in both conditions (delta)S is also positive, indicating an increase in entropy.  For the most part any system will always trend toward an insrease in entropy.  \n\nThe question then arises, what happens when (delta)S is negative (decreased entropy) and (delta)H is negative.  For most situations T(delta)S will be large enough to create a positive (delta)G from our equation.  In very rare cases will (delta)H - T(delta)S result in a negative answer.  However, the equation shows this as a possibility when (delta)G= large negative (delta)H - T x small negative (delta)S => large negative # minus a smaller negative => larger negative + smaller # = negative.  (Hope you follow my train of thought).  Therefore in the right situation - (delta) can be negative and bonding/reaction can occur when entropy decreases.  Eliminate E and C is the correct answer.\n\nCareful what you ask for.  Wikipedia site looks pretty informative but very detailed and you have to use a few different pages.
878075
First of all let's see the expression of Kp for the first equation\n\nH2+Br2 <=> 2HBr\n\nKp=P(HBr)^2 / (P(H2)*P(Br2))= 3.5*10^4\n\na) HBr <=>1/2H2+ 1/2Br2\n\nKp1=[P(H2)^1/2]*[P(Br2)^1/2] / P(HBr)\nBut if you look carefully  Kp1=1/squareroot(Kp) =\n=1/(SQRT(3.5*10^4))= 5.35*10^-3\n\nRemember that since squareroot of x= x^1/2\nKp1=[P(H2)^1/2]*[P(Br2)^1/2] / P(HBr)=\n\n= [P(H2)*P(Br2)/(P(HBr)^2)] ^(1/2)=\n\n=1 / [P(HBr)^2/(P(H2)*P(Br2))] ^(1/2) =\n\n=(1/Kp)^(1/2) = 1/Kp^(1/2)=1/Squareroot(Kp)\n\nI am sorry I couldn't write it more clearly but it is a limitation of the site and we can't write equations in a better way.\n\nb)2HBr <=> H2+ Br2\nIt is reverse of the initial reaction so\n\nKp2=(P(H2)*P(Br2)) / P(HBr)^2= 1/Kp= 1/(3.5*10^4)= 2.86*10^-5\n\nc)1/2H2+1/2Br2 <=>HBr\n\nis the reverse of a do Kp3=1/Kp1= 1/(5.35*10^-3)=1.87*10^2
673440
The activation energy of a reaction is different for every reaction and is totaly independant of wether the reaction is exothermic or endothermic. For any reaction to proceed the minimum energy must be equal to or higher than the activation energy.\n\nNow where your question can come in is, i think, is if the reaction is reversible. In one direction the reaction will be exothermic and in the opposite direction the reaction will be endothermic. For both the forward and backward reactions the activation energy is the same and there is nothing that can be done to change that.
793972
A exothermic\nB both are favoured\nC 1 equilibrium shifts towards the right\n    2 equilibrium shifts towards left\n    3 equilibrium shifts towards left\n    4 equilibrium shifts towards right\n    5 no change
419731
yes:\n\nThe first answer is in the right direction.\n\nA reaction engine is one that moves in a direction opposite to the thrust of combustion.\n\nA reaction engine works on the basis of Newton's third law of motion: \n\n"For every action there is an equal and opposite reaction."\n\nIn a reaction engine fuels are burnt and the extra pressure due to formation of more volume of gases is directed to get out of a nozzle. This forms "action"\n\nAs a result of action the engine or the body containing the system moves in opposite direction.\n\nRockets and Jet Planes are the examples of reaction engine.\n\nThe movement of a balloon as a result of escaping gas from an opening is one very simple example.
799142
Hi Kitty,\nHere is what, I think, you need.\n\nExothermic, dH < 0, Entropy increases (dS > 0), Reaction is spontaneous and dG < 0\n\nExothermic, dH < 0, Entropy decreases (dS < 0), Reaction is spontaneous only at low temps, if |T dS| < |dH|\n\nEndothermic, dH > 0, Entropy increases (dS > 0), Reaction is spontaneous only  at high temps, if T dS > dH\n\nEndothermic, dH > 0, Entropy decreases (dS < 0), Reaction is not spontaneous, dG > 0
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
766444
time to the minus one.\n\nDo a dimensional analaysis on a rate equation to confirm this.