Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The ΔHf of C2H4 is 52.26 kJ/mol, CO2 is -393.509 kJ/mol, and H2O is -241.818 kJ.) C2H4 (g) + 3O2(g) 2CO2 (g) + 2H2O(g)

ΔHrxn = (-345.64 kJ, -583.07 kJ, or -1,322.91 kJ).

The reaction is: (Endothermic or Exothermic).

We can construct an enthalpy diagram for it:

(please take a look at the picture).

-write the equation, indicate the elements that form the reactants/products. The enthalpy change for formation of elements itself is always 0, so nothing needs to be indicated for O2(g).  

*make sure you multiply the enthalpy with the no. of moles for reactants/products with coefficients more than 1 in the equation.

Clockwise arrows = anti-clockwise arrows.

Therefore, by Hess's law,

ΔHrxn + ΔHf [C2H4(g)] = ΔHf [CO2(g)] + ΔHf [H2O(g)]

ΔHrxn + 52.26 = 2 (-393.509) + 2 ( -241.818)

ΔHrxn = -1,322.91 kJ/mol

Since the enthalpy change is negative, the reaction is exothermic.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol