Measurements show that enthalpy of a mixture of gaseous reactants decreases by 228. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -55kJ of work is done on the mixture during the reaction.

Calculate the change in energy of the gas mixture during the reaction.

Is the reaction exothermic or endothermic.

Ok here's what I got:

(Delta)H = -288. kJ
(Delta)P = 0 because it's constant
W = -55 kJ

Also, I know that the reaction is exothermic because deltaH is negative, meaning the su

Question

Measurements show that enthalpy of a mixture of gaseous reactants decreases by 228. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -55kJ of work is done on the mixture during the reaction.

Calculate the change in energy of the gas mixture during the reaction.

Is the reaction exothermic or endothermic.

Ok here's what I got:

(Delta)H = -288. kJ
(Delta)P = 0 because it's constant
W = -55 kJ

Also, I know that the reaction is exothermic because deltaH is negative, meaning the su

anonymous · Answer

Meaning the substance is releasing heat to its surroundings.

I just don't know how to use the equations:

W=-P(delta)v
DeltaH=deltaE+delta(Pv)

aaronq · Answer

i think they want internal energy

$\sf U=q+w$

q= H when P is constant

$\sf \Delta U=- 228~ kJ -55~kJ$

anonymous · Answer

Of course! Thank you!

So then whats the difference between ∆H and ∆E?

aaronq · Answer

no problem. look at this link http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/chemical.php

anonymous · Answer

much obliged