Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
X moles of CO₂, and x moles of H₂O
CxH2x + 3x/2 O₂ -------> xCO₂ + xH₂O
CxH2x or CnH2n ----> ALKENE ( hydrocarbon with double bond)
Answer:
Nitrogen
Explanation:
Nitrogen present 78% in the earth's atmosphere
The chemical equation would be:
2NO(g) + O2(g) --> 2NO2 (g)
<span>At equilibrium state, the partial pressure of the gases would be as follows : </span>
<span>NO = 522 - 2x </span>
<span>O2 = 421 - x </span>
<span>NO2 = 2x </span>
<span>- - - - - - - - - - - - -</span>
<span>943 - x = 748 </span>
<span>x = 195</span>
Calculating for Kp,
<span>Kp = (NO2)^2/ ((NO)^2 * (O2)) </span>
<span>Kp = (2 * 195)^2/ ((522 - 2 * 195)^2 * (421 - 195)) </span>
<span>Kp = 0.0386 </span>
