Enthelpy Change of Combustion.

  • Thread starter Thread starter jinhuit95
  • Start date Start date
  • Tags Tags
    Change Combustion
Click For Summary
SUMMARY

The discussion focuses on calculating the enthalpy change of combustion for ethanol based on a specific experiment. When 1.00g of ethanol is burned, it heats 100g of water from 15°C to 65°C with a process efficiency of 70%. The correct calculation for the heat absorbed by water is 14630J, leading to an enthalpy change of -673 kJ mol^-1. The error identified in the calculation was the incorrect application of efficiency; it should be divided by 70% instead of multiplied.

PREREQUISITES
  • Understanding of thermodynamics and enthalpy changes
  • Familiarity with the specific heat capacity formula (q = mcΔT)
  • Knowledge of calculating moles from mass and molar mass
  • Basic proficiency in unit conversions and efficiency calculations
NEXT STEPS
  • Review the concept of enthalpy change in combustion reactions
  • Learn about the specific heat capacity of various substances
  • Study the principles of efficiency in thermodynamic processes
  • Explore common mistakes in stoichiometric calculations in chemistry
USEFUL FOR

Chemistry students, educators, and anyone involved in thermodynamics or combustion analysis will benefit from this discussion.

jinhuit95
Messages
28
Reaction score
0

Homework Statement


When 1.00g of ethanol was burned under a container of water, it was found that 100g of water was heated from, 15°C to 65°C, The process was known to be only 70% efficient. Calculate the enthalpy change of combustion of ethanol taking specific heat capacity of water to be 4.18 J g^-1 k^-1.


Homework Equations


q= mc ΔT; ΔHr= - q/ moles of limiting reagent.


The Attempt at a Solution


I calculate q = 100 x 4.18 x 50 x(70/100) = 14630J.
Then I found the number of moles of ethanol which is 1/46 = 0.0217
Therefore enthalpy change will then be -14630/0.0217 = -672980J = -673 kJ mol^-1 but the answer given is -1.37 x 10^3 kJ mol^-1. Where did I go wrong??
 
Physics news on Phys.org
In your first step you should be dividing by 70% rather than multiplying by 70%.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 19 ·
Replies
19
Views
4K