want omited. There is an extra credit question at the end. R 82.06 (atm cm3)/(mo
ID: 553279 • Letter: W
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want omited. There is an extra credit question at the end. R 82.06 (atm cm3)/(mol K) R 0.08206 (atm L)/(mol K) R 8.314 (Kg m2 (s mol K) P = nRT/(V-nb)-a(n/V)2 -= (3RT/Mw)1/2 760 torr = 760 mmHg 1 atm specific heat of water = 4.18 J/(g oC) PV = nRT q = m-SH-AT Xa = Pa/P,otal-na/ntotal 1. (5 points) From the information given below calculate the enthalpy change for the reaction NH3 (g) + HCI(g) NH4Cl(s) Given: HCI(g) 1 /2 H2 (g) + 1/2 Cl2 (s) 4H2(g) + N2(g) + C12(s) 2 NH4C1(s) N2(g) + 3H2 (g) 2 NH3(g) 1° =-92.5 kJ AH -92.2 kJ a) -7,890 kJ b) -360 kJ c) -1,160 kJ d) +13,284 kJExplanation / Answer
You just need to rearrange the equations, you need to let the reactants on the left and products on the right
from my perspective the first step should be tu reverse the third equation
2 NH3 ====== N2 + 3H2
let´s write the equations
HCl ===== 1/2 H2 + 1/2 Cl2
4H2 + N2 + Cl2 ======= 2 NH4Cl
2 NH3 ====== N2 + 3H2
The reactanst and the products are where we want them but if we summarize all the hydrogens and chlorides they are not balanced also we have 2 NH3, 2 NH4Cl and 1 HCl, the best thing to do is to multiply by 2 equation 1
HCl ===== 1/2 H2 + 1/2 Cl2 * 2
2HCl ===== H2 + Cl2; let´s summarize the equations:
-----------------------------------------------------------------------------------------
2 HCl ===== H2 + Cl2
4H2 + N2 + Cl2 ======= 2 NH4Cl
2 NH3 ====== N2 + 3H2
___________________________________
2 HCl + 2 NH3 ======= 2NH4Cl
the values must be changed, for the first equation we take the enthalpy and multiply it by 2
-92.5 * 2 = -185
the third enthalpy with a value of -92.2 turns positive to 92.2,this is because we reversed the equation
total enthalpy is = -185 + 92.2 + (-628) = -720.8 KJ, but this is the enthalpy for a reaction of
2 HCl + 2 NH3 ======= 2NH4Cl, we need to divide the enthalpy by 2 in order to get the value from the equation we want.
-720.8 / 2 = -360.4 KJ
Anwer is b)
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