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dem82 [27]
3 years ago
9

When ch4(g) reacts with h2o(g) to form h2(g) and co(g), 206 kj of energy are absorbed for each mole of ch4(g) that reacts. write

a balanced thermochemical equation for the reaction with an energy term in kj as part of the equation. note that the answer box for the energy term is case sensitive. use the smallest integer coefficients possible and put the energy term in the last box on the appropriate side of the equation. if a box is not needed, leave it blank. + + + +?
Chemistry
1 answer:
jeka57 [31]3 years ago
7 0
1) Chemical reaction: CH₄ + H₂O → 3H₂ + CO ΔH = + 206 kJ/mol.
2) Chemical reaction: CO + H₂O → CO₂ + H₂, ΔH = + 2,8 kJ/mol.
3) Chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = -802 kJ/mol.
4) Chemical reaction: 2H₂ + O₂ → 2H₂O ΔH = 2·(-242 kJ7mol) = - 484 kJ/mol.
Endothermic reaction (ΔH>o) and exothermic reaction (ΔH<span><0).</span>
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What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
Marina86 [1]

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

P = 1.75 atm

n = 2.5 moles

T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

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A negative △H means an exothermic reaction.<br><br> True<br> False
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ammonium nitrate, which is prepared from nitric acid, is used as a nitrogen fertilizer. determine the percent composition of amm
Alina [70]

Answer:

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5 0
2 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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