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This is one of our many quizzes on Enthalpy. Click on this to access more such competitive quizzes on Thermodynamics and Thermochemistry.

Enthalpy Quiz 2

chemistry

 If the enthalpy of combustion of ethanol is $-1366.95kJ/mol$, what will be the amount of heat produced at NTP?

 The combustion of ethanol takes place as following $ \\ {{C}_{2}}{{H}_{5}}OH(g)+3{{O}_{2}}(g)\to 2C{{O}_{2}}(g)+3{{H}_{2}}O(l)$ 

 $\Delta E=\Delta H-P\Delta V\\ \text{But }P\Delta V=\Delta {{n}_{g}}RT \\ R=8.314 J/mol.K = \dfrac{8.314}{1000} kJ/mol.K; \\ \Delta {{n}_{g}}=-1; \\ \text{ and } T = 25{}^\circ C = 298 K\\ \text{So,}\Delta E=\Delta H-\Delta {{n}_{g}}RT \\ \Delta E=-1366.95- \left(-1\times \dfrac{8.314}{1000}\times 289\right) \\ \Delta E=-1364.47\text{ }kJ/mol \\ $ Hence the amount of heat produced will be $1364.47 kJ/mol$

 How many liters of water are vaporized at $1$ bar and $100{}^\circ C$ if the change in internal energy is $10.041 kJ/mol$? $($molar enthalpy change, $\Delta H=41kJ/mol; density = 1g/L)$

  The reaction of the change is as follows $ \\ {{H}_{2}}O(l)\xrightarrow{\Delta }{{H}_{2}}O(g)$ 

 We know that $ \Delta U=\Delta H-\Delta {{n}_{g}}RT.\\ \text{ So, }\Delta {{n}_{g}}=\dfrac{\Delta H-\Delta U}{RT} \\ \Delta {{n}_{g}}=\dfrac{41-10.041}{8.3\times 373}=0.01\text{ moles}  \\ $ $0.01$ moles of water $= 0.18g$ of water $= 0.18L$ of water.

  3900kJ of heat is required to raise the temperature of 75kg of plastic. The change in temperature is $(c=2.60 kJ/(kg.{}^\circ C))$

 The formula of specific heat capacity is  $ \\ c=\dfrac{Q}{\Delta T\times m}$ 

 Substituting the values in the specific heat capacity formula we get $\\ 3900=mc\Delta T \\ \Delta T=\dfrac{3900}{75\times 2.60} \\ \Delta T=20{}^\circ C \\ $

  The temperature dependence of equilibrium constant K for a reaction is given by the graph given below.  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9225663d-446b-423d-bf37-3046ec5adbc01386390484671805474.png"/> $ \\ $ It can be concluded that the reaction is 

 The relation between the equilibrium constant $K$ and temperature can be derived from the Gibbs free energy and the equilibrium equation $\Delta G{}^\circ =-RT\ln K$ 

  From the equations $\Delta G{}^\circ =-RT\ln K$ and the equation $\Delta G{}^\circ =\Delta H{}^\circ -T\Delta S{}^\circ$ we can say that $ \\ \ln K=\dfrac{-\Delta H{}^\circ }{RT}+\dfrac{\Delta S{}^\circ }{R}\\ $ If a graph is plotted between$ lnK$ and $\dfrac{1}{T}, $ the slope would be given by $\dfrac{-\Delta H{}^\circ }{R} \\ $ Since the slope is positive, $\Delta H{}^\circ $ will be negative, hence the reaction is exothermic.

 Enthalpy change upon complete combustion of $64.4g$ of carbon monoxide is: $ \\ CO(g)+\dfrac{1}{2}{{O}_{2}}(g)\to C{{O}_{2}}(g) \\ \Delta H=-280kJ$ 

 The molar mass of carbon monoxide is $28g/mol$.

Moles of $CO$ in $64.4g = \dfrac{64.4}{28}=2.3$  $ \\ $ The energy released upon combustion of $1$ mole of $CO$ is $280kJ$. $ \\ $ The energy that will be released upon combustion of $2.3$ moles of $CO$ is $2.3\times 280= 644 kJ$. $ \\ $ Since combustion is exothermic energy, enthalpy change will be negative.