Calculate δhrxn for the following reaction: ch4(g)+2o2(g)→co2(g)+2h2o(l) use the following reactions and given δh values. ch4(g)+o2(g)→ch2o(g)+h2o(g), δh=−284 kj ch2o(g)+o2(g)→co2(g)+h2o(g), δh=−527 kj h2o(l)→h2o(g), δh= 44.0 kj

Answer :

-899Kj

Explanation

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= {ΔHH2O(l) + ΔHCO2} - ΔHCH4

We do not put that of Molecular oxygen in the equation as it is zero

ΔHH2O(g) - ΔHH2O(l) = 44

-527 = {ΔHH2O(g) + ΔHCO2} - ΔH{CH2O}

-284 = {ΔHCH2O + ΔHH2O(g) } -ΔH{CH4}

Adding the last 2 equations together will yield the following:

-527-284 = 2ΔH{H2O}(g) + ΔHCO2 - ΔHCH4

From the second equation,

ΔHH2O(g) = 44 + ΔHH2O(l)

Substitute this in the added equation

-811 = 2(44 + ΔHH2O(l)) + ΔHCO2 - ΔHCH4

-811 = 88 + 2ΔHH2O(l) + ΔHCO2 - ΔHCH4

-899 = 2ΔHH2O(l) + ΔHCO2 - ΔHCH4

The heat of reaction is thus -899kJ

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

i think so it’s will be b yes i believe its