If the standard enthalpy change $\left(\Delta_{r} H^\theta\right)$ for the reaction $H_{2(g)} + Br_{2(l)} \rightarrow 2 HBr_{(g)}$ is $-72.8 \ kJ$,the standard enthalpy of formation $\left(\Delta_{f} H^\theta\right)$ of $HBr_{(g)}$ (in $kJ \ mol^{-1}$) is

The heat change for the reaction $H_2 + \frac{1}{2} O_2 \to H_2 O$ is called

Which of the following has standard enthalpy of formation equal to zero?

The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?

At $298 \ K$, the enthalpy change (in $kJ$) for the reaction given below is: $CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$
Given:
$1) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H^{\ominus} = -286 \ kJ$
$2) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H^{\ominus} = -394 \ kJ$
$3) \ CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} ; \Delta H^{\ominus} = -890 \ kJ$

If the heat of neutralization of an acid-base reaction is $56 \ kJ \ mol^{-1}$,then the substances could be: