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anonymous
 4 years ago
How do you balance
C2H2 + O2 > CO2 + H2O??
Is it C2H2 + 5/2 O2 > 2CO2 + H2O
or 2C2H2 + 5O2 > 4CO2 + 2H2O ?
anonymous
 4 years ago
How do you balance C2H2 + O2 > CO2 + H2O?? Is it C2H2 + 5/2 O2 > 2CO2 + H2O or 2C2H2 + 5O2 > 4CO2 + 2H2O ?

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Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1All coefficients must be whole numbers. Often, substeps will include fractional coefficients, but the final answer must be only whole numbers.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 C2H2 + 5 O2 > 4 CO2 + 2 H2O

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1To answer your question explicitly, your second answer is correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's what i think too, but my teacher put C2H2 + 5/2 O2 > 2CO2 + H2O The standard enthalpy of combustion of C2H2 is 1301kJ. the standard enthalpies of formation of CO2 and H2O are 394 and 286kJ, what is delta H f for C2H2?? so in this case, u will still use the 2nd equation?

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1It shouldn't matter, I don't believe. In either case, you would end up multiplying the enthalpy changes by the coefficients anyway. However, I'm not 100% sure on this. Let's check. For example, in this case it would be something like:\[\Delta H_{rxn}^0=\sum r \Delta H_f^0(products)\sum r \Delta H_f^0(reactants)\]Where r is the coefficients of the compounds. So, using the wholenumber reaction equation:\[\Delta H_{rxn}^0= [(4)(394kJ)+(2)(286kJ)][(2)(1301kJ)+5(0kJ)]\]\[\Delta H_{rxn}^0=454.kJ\]Now, if we're correct, the reduced reaction equation should produce the same answer. Let's try:\[\Delta H_{rxn}^0=[(2)(394kJ)+(1)(286kJ)][(1)(1301kJ)+(5/2)(0kJ)]\]\[H_{rxn}^0=227.kJ\] Well, based on that, it would seem that the chemical formula obviously does matter. And I just remembered why this is so.

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1I believe that the formula for the enthalpy of formation for a compound must be written in such a way that the coefficient of the compound you are trying to find the enthalpy of formation for is 1.

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1Once again, though, I'm not sure. This should be verified by someone who knows.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh ok thank you so much :) I will check back the notes again!

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1Are there any more equations involved in this specific problem?

Rogue
 4 years ago
Best ResponseYou've already chosen the best response.2It does not matter which way you solve for it. If you were to use the the delta H rxn of the balanced reaction, you would be getting the enthalpy of formation for 2 moles of C2H2; you can just halve that to get the standard enthalpy of formation which is for 1 mole of the molecule/compound. Solving for it using the reaction with the .5 mol O2 will directly give the you standard enthalpy of formation. So yeah, Xishem is right. The standard enthalpy of formation is only for 1 mole, but it can solved for via many different ways.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have many enthalpy ques like that but yea, i know why juz ask rogue said, it is 2mol in here, so all i have to do is to divide it to get the answer cuz all my other questions only have 1 mol of that but i remember only for this kind of equation can have fraction in the equation, and there is a type of reaction that it must not have fraction of coeeficient , so i messed up those 2.. i need to find out which one is which.. and thanks for both of ur help again!

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.1I apologize for not having a full understanding of the subject. I hope I didn't confuse you further.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem :) I understand it better now !
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