Calculate the enthalpy change, delta H, for CH4(g) + 4Cl2 (g) rightarrow CCl4 (I

Question

Calculate the enthalpy change, delta H, for CH4(g) + 4Cl2 (g) rightarrow CCl4 (I) + 4HCI(g) from the following thermochemical data C(graphite) + 2H2 (g) rightarrow CH4 (g); delta H = -74.9kJ C(graphite) + 2Cl2 (g) rightarrow CCl4(l); delta H = -139.()kJ 1/2 H2 (g) + 1/2 Cl2 (g) rightarrow HCI (g); delta H = -92.3kJ Is the reaction endothermic or exothermic? Calculate the enthalpy change, delta H, for the reaction 2C(graphite) + 3H2(g) rightarrow C2H6(g) from the following thermochemical data C(graphite) + O2 (g) rightarrow CO2(g); delta H = -393.5kJ H2 (g) + 1/2O2(g) rightarrow H2O(1); delta H = -285.8kJ 2C2H6(g) + 702 (g) rightarrow 4CO2(g) + 6H20(l); delta H = -3119.6kJ Using AHt degree values, calculate the enthalpy change delta H degree for the combustion of glucose, C6H12O6 + 6O2(g) rightarrow 6CO2(g) + 6H2O(I) delta H5 degree -1273kJ -393.509kJ -285.83 How many kJ are produced when 1.00 grams glucose is burned? Using delta Ht degree values, calculate the enthalpy change delta H degree for the combustion of glucose, C2H5OH(I) + 3O2(g) rightarrow 2CO2(g) + 3H2O(l) delta Ht degree -277.6U -393.509kJ -285.83 Calculate the amount of heat given off when 5.36 grams of ammonia read completely in the reaction below 2NH3 (g) + 3O2 (g) +2CH4(g) rightarrow 2HCN(g) + 6H2O(g); delta H = -939kJ How much heal is evolved when 2.38grams of H2O are produced in the reaction above?

Explanation / Answer

1) if we name given equations as 1 , 2 , 3 then we get required equation

-eq(1) + eq(2) + 4eq(3)
dH = -(-74.9) + (-139) +4(-92.3) = -433.3 KJ

reaction is exothermic

2) if given reactions named 1,2,3 then we get required eq

2eq(1) + 3eq(2) -1/2eq(3)

dH = 2( -393.5)+3(-285.8)-1/2(-3119.6)

= -84.6 KJ

3) dH = 6(-393.5) +6(-285.8) -(-1273) = -2802.8 KJ

when 1gm glucose = (1/180) moles burnt dH = ( -2802.8/180) = -15.57   , hence 15.57 KJ heat is produced

4) dH = 2( -393.5)+3(-285.8) -(-277.6) = -1366.8 KJ

5) 5.36 gm NH3 = 5.36/17 = 0.3153 moles

dH = 0.3153 x (-939/2) = -148 KJ

2.38 gm H2O = 2.38/18 = 0.132 moles H2O

dH = 0.132 x (-939/6) = -20.69 KJ

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.