Given the propane molecule C3H8(g)! a) What is the balanced equation for the combustion of propane in O2?(20 pts.) b) What is the standard enthalpy change ΔH0 for the reaction at 25°C? (20 pts.) c) Finally, is this reaction exothermic or endothermic, under what condition is the answer to part b) the heat of the reaction? (10 pts.)

The answer to your question is below

Explanation:

Propane = C₃H₈

Process

1.- Write the chemical reaction

                     C₃H₈  +  O₂   ⇒   CO₂  +  H₂O

balanced chemical reaction

                      C₃H₈  +  5O₂   ⇒  3CO₂  +  4H₂O

               Reactants      Elements     Products

                      3                    C                  3

                      8                    H                  8

                     10                    O                 10

b) Standard enthalpy

Propane                  -104.7 kJ/mol

Oxygen                         0   kJ/mol

Carbon dioxide      -393.5 kJ/mol

Water                      -241.8 kJ/mol

ΔH° = ∑ΔH° products - ∑H° reactants

ΔH° = 3(-393.5) + 4(-241.8) - [(-104.7) + 0]

-Simplification

ΔH° = -1180.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

ΔH° = - 2042.5 kJ/mol

This reaction is exothermic because ΔH° is negative.

away from the curb

explanation:

the wheels would then hit the curb if it started to rolling, decreasing the chance of an accident.

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