10 kmol of methane gas are heated from 1 atm and 298 K to 1 atm and 1000 K. Calculate the total

10 kmol of methane gas are heated from 1 atm and 298 K to 1...

$10 \mathrm{kmol}$ of methane gas are heated from 1 atm and $298 \mathrm{K}$ to $1 \mathrm{atm}$ and $1000 \mathrm{K}$. Calculate the total amount of heat transfer required when $(a)$ disassociation is neglected and $(b)$ when disassociation is considered. The natural logarithm of the equilibrium constant for the reaction $\mathrm{C}+2 \mathrm{H}_{2}$ $\rightleftharpoons \mathrm{CH}_{4}$ at $1000 \mathrm{K}$ is $2.328 .$ For the solution of part (a) use empirical coefficients of Table $A-2 c$. For the solution of part ( $b$ ) use constant specific heats and take the constantvolume specific heats of methane, hydrogen and carbon at $1000 \mathrm{K}$ to be $63.3,21.7,$ and $0.711 \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K},$ respectively. The constant-volume specific heat of methane at $298 \mathrm{K}$ is $27.8 \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K}$.

Key Concepts

Sensible Heat Transfer

This concept involves calculating the heat required to raise the temperature of a substance without changing its phase or chemical structure. It is done by integrating the heat capacity (which may depend on temperature) over the temperature range of interest. In thermodynamics, sensible heat calculations are essential when the heating process is purely physical, and they form the basis for understanding energy input in both non-reactive and reactive processes.

Temperature?Dependent Specific Heat Capacity

For accurate computations, especially over a wide temperature range, the specific heat capacity is often expressed as a function of temperature. Empirical coefficients (such as those in Table A-2c) are used with polynomial forms to calculate the heat capacity at any particular temperature. This method allows for more precise integration and energy calculation compared to using a single average or constant value over the range.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the forward and reverse reaction rates are equal, resulting in constant concentrations of all species involved. In the context of heating methane, chemical equilibrium concepts are invoked when dissociation occurs at high temperatures. The equilibrium constant (or its logarithm) quantifies the position of equilibrium and is used to determine the extent to which methane dissociates into its constituent elements or reaction products as temperature changes.

Reaction Enthalpy and Effects of Dissociation

When a chemical reaction occurs during the heating process, additional energy is required or released due to the reaction enthalpy. In cases where dissociation is considered, the energy balance must account not only for the sensible heat needed to raise the temperature but also the energy associated with breaking and forming chemical bonds. This makes the analysis more complex as it involves both thermal and chemical energy changes within the system.

Assumption of Constant Specific Heats

In some scenarios, especially when high accuracy is not critical or when temperature variations are moderate, specific heats are assumed to be constant. This simplification aids in analytical calculations. In the provided problem, constant-volume specific heats are assumed for the species involved during the high-temperature part of the process, streamlining the calculation of heat required under reactive conditions.

Transcript

00:01 All right guys, the heat transfer in this case is determined from the simple energy balance on the methane.

00:12 So q is equal to n into h2 bar minus h1 bar minus ru into t2 minus t1.

00:26 So we have n is equal to integral from range 1 to 2 of cpdd minus r u t2 minus t1.

00:48 So we have n into a minus r u into t2 minus t 1 plus b over 2.

01:09 Into t2 square minus t 1 plus c over 3 into t 2 cube minus t 1 plus d over 4 into t 2 raise to power 4 minus t 1 raise to power 4 so now we put the values and we get 10 into 19 .89 minus 8 .314 into 1000 minus 298 plus 5 .024 into 10 raised to power minus 2 divided by 2 multiplied by 1000 square minus 298 square and plus 1 .269 9 to 6 .969 raise to power minus into 10 raised to power minus 5 divided by 3 multiplied by 1 ,000 cube minus 298 q minus 11 .01 into 10 raised to power minus 9 divided by 4 into 1 ,000 raised to power 4 minus 2 98 raised to power 4 kilojoules and we get 324 -0 kilojowl.

03:14 Now, guys, moving ahead, and now we solve part b of this question, and that is that the stociometric reaction for this example, and the corresponding coefficients are c -h -4 is equal to c plus 2h2, so v -ch -4 is equal to 1 vc is equal to 1 v -c is equal to 1 v h2 is equal to 2 the actual reaction with the given coefficients b and c that are expressed through term a k r c h4 is equal to a c h 4 plus b c h2 so we have the values b is equal to 1 minus a and c is equal to 2 minus 2a so the coefficients are determined from the equilibrium uh coefficient relations and here we have e raised to power minus natural log kp is equal to nc vc into n hc v sorry n h2 raised to power vh2 divided by n c h 4 raised to power v c h 4 into p divided by n raised to power v h 2 minus vch 4 now we have the value e raised to power minus 2 .328 is equal to 1 minus a into 2 .328 is equal to 1 minus a into 3 minus 2a square divided by a into 1 divided by 3 minus 2a raised to power 1 plus 2 minus 1.

05:27 So we have the equation minus 4 .39 a cube plus 13 .17a square minus 12 .88 a plus 4 is equal to 0.

05:51 So from here we calculate the values of a, that is a is equal to 6403.

06:03 B is equal to 0 .3597...

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