(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =

=
= 3.05 x 10¹⁸
Every mole of CH4 used, three moles of H2 are produced, so 2 moles of CH4, would be 6 moles of H2 produced
Answer:
Explanation:
13 ) symbol of enthalpy change = Δ H .
14 ) enthalpy change is nothing but heat absorbed or evolved .
During fusion enthalpy change
Δ H .= m Lf , m is mass and Ls is latent heat of fusion
During evaporation, enthalpy change
Δ H .= m Lv , m is mass and Lv is latent heat of evaporation
during temperature rise , enthalpy change
Δ H = m c Δ T
In case of gas , enthalpy change can be calculated by the following relation
Δ H = Δ E + W , Δ E is change in internal energy , W is work done by gas.
15 ) When enthalpy change is negative , that means heat is released to the environment .So reaction is called exothermic .
when heat is absorbed enthalpy change is positive . Reaction is endothermic.
Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C = 
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
![K = \frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)
where
K is given as ; 78.2 atm-1.
So, we have:
![78.2=\frac{[0.0530-x]}{[x][x]}](https://tex.z-dn.net/?f=78.2%3D%5Cfrac%7B%5B0.0530-x%5D%7D%7B%5Bx%5D%5Bx%5D%7D)


Using quadratic formula;

where; a = 78.2 ; b = 1 ; c= - 0.0530
=
or 
=
or 
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:

where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???

V = 1.64604
V ≅ 1.65 L
Answer:24.31
Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97
Contribution made by isotope of mass 24.99=24.99×10.00=249.9
Contribution made by isotope of mass 25.98=25.98×11.01=286.04
Total contribution=1894.97+249.9+286.04=2430.91
Average mass=2430.91÷100
=24.31